NCERT Solutions Class 12 Physics Chapter 11

Q.1 Find the
(a) Maximum frequency, and
(b) Minimum wavelength of X-rays produced by 30 kV electrons.

Ans.

$\begin{array}{l}{\text{(a) \hspace{0.33em}As maximum kinetic energy of photoelectrons is given by K}}_{\text{max}}{\text{= eV}}_{\text{o}}\text{,}\\ {\text{where V}}_{\text{o}}\text{\hspace{0.33em}is the stopping potential of anode.}\\ {\text{Thus using\hspace{0.33em} K}}_{\text{max}}{\text{=\hspace{0.33em}\hspace{0.33em}eV}}_{\text{ο}}\\ {\text{= hν}}_{\text{max}}\\ {\text{we get\hspace{0.33em}ν}}_{\text{max}}\text{=\hspace{0.33em}\hspace{0.33em}}\frac{{\text{eV}}_{\text{ο}}}{\text{h}}\\ \text{Where}\\ {\text{h = 6.63 ×\hspace{0.33em}10}}^{\text{-34}}\text{Js;\hspace{0.33em}}\\ {\text{e = 1.6 ×\hspace{0.33em}10}}^{\text{-19}}\text{\hspace{0.17em}C}\\ \text{Thus}\\ {\text{ν}}_{\text{max}}\text{=}\frac{\left({\text{1.6 × 10}}^{\text{-19}}\text{C × 30,000\hspace{0.33em}V}\right)}{{\text{6.63 × 10}}^{\text{-34}}\text{Js\hspace{0.33em}}}\\ {\text{ν}}_{\text{max}}{\text{= 7.24 × 10}}^{\text{18}}\text{\hspace{0.17em}Hz\hspace{0.17em}}\\ {\text{(b)\hspace{0.33em}As\hspace{0.33em}\hspace{0.33em}λ}}_{\text{min}}\text{=}\frac{\text{hc}}{\text{eV}}\\ \text{=}\frac{\text{c}}{{\text{ν}}_{\text{max}}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{{\text{7.24 × 10}}^{\text{18}}\text{\hspace{0.33em}Hz}}\\ {\text{λ}}_{\text{min}}{\text{= 0.414 × 10}}^{\text{-10}}\text{\hspace{0.33em}m}\\ \text{= 0.041 nm}\\ \text{Hence,\hspace{0.33em}the\hspace{0.33em}minimum\hspace{0.33em}wavelength\hspace{0.33em}of\hspace{0.33em}X-rays\hspace{0.33em}produced}\\ \text{is\hspace{0.33em}0.041 nm\hspace{0.17em}.}\end{array}$

Q.2 The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the(a) Maximum kinetic energy of the emitted electrons,(b) Stopping potential, and(c) Maximum speed of the emitted photoelectrons?

Ans.

$\begin{array}{l}\text{(a)}\\ \text{Work function,}\\ \mathrm{\varphi }\text{\hspace{0.33em}= 2.14 \hspace{0.17em}eV}\\ {\text{= 2.14 × 1.6 × 10}}^{\text{-19}}\text{J}\\ {\text{= 3.424 × 10}}^{\text{-19}}\text{J}\\ \mathrm{\nu }{\text{= 6 × 10}}^{\text{14}}\text{Hz}\\ {\text{h = 6.26 × 10}}^{\text{-34}}\text{Js}\\ \text{As per Einstein’s photoelectric equation,\hspace{0.33em}maximum}\\ \text{kinetic energy\hspace{0.33em}of photoelectrons is}\\ \text{K.E = h}\mathrm{\nu }\text{–}\mathrm{\varphi }\\ \text{K.E =}\left({\text{6.626 × 10}}^{\text{-34}}{\text{Js × 6 × 10}}^{\text{14}}\text{\hspace{0.33em}Hz\hspace{0.33em}}\right)\text{–}\left({\text{3.424 × 10}}^{\text{-19}}\text{\hspace{0.33em}J}\right)\\ \text{K.E =}\left(\text{3.976 – 3.424}\right){\text{× 10}}^{\text{-19}}\text{\hspace{0.33em}J}\\ {\text{= 0.552 × 10}}^{\text{-19}}\text{J}\\ \text{Thus\hspace{0.33em}K.E\hspace{0.33em}in\hspace{0.33em}eV\hspace{0.33em}will\hspace{0.33em}be\hspace{0.33em}}\frac{{\text{0.552 x 10}}^{\text{-19}}\text{\hspace{0.33em}J}}{{\text{1.6 x 10}}^{\text{-19}}\text{\hspace{0.33em}C}}\text{= 0.345 eV}\\ \text{= 0.35 eV}\\ \text{(b)}\\ \text{By\hspace{0.33em}the\hspace{0.33em}relation}\\ {\text{eV}}_{\text{o}}\text{= Maximum\hspace{0.33em}K.E}\\ {\text{eV}}_{\text{o}}\text{= 0.35 eV}\\ {\text{V}}_{\text{o}}\text{= 0.35 V\hspace{0.17em}}\\ \text{(c)}\\ {\text{As,\hspace{0.33em}K.E}}_{\text{max}}\text{=}\frac{\text{1}}{\text{2}}{{\text{mv}}^{\text{2}}}_{\text{max}}\\ \text{= 0.35 eV}\\ {\text{= 0.552 × 10}}^{\text{-19}}\text{J}\\ {{\text{Thus\hspace{0.33em}v}}^{\text{2}}}_{\text{max}}\text{=}\frac{\left({\text{0.552 × 10}}^{\text{-19}}\text{\hspace{0.33em}J}\right)\text{× 2}}{\text{m}}\\ \text{=}\frac{\left({\text{0.552 × 10}}^{\text{-19}}\text{\hspace{0.33em}J}\right)\text{× 2}}{{\text{9.1 × 10}}^{\text{-31}}\text{\hspace{0.33em}kg}}\\ {\text{v}}_{\text{max}}\text{=}\sqrt{{\text{0.121318 × 10}}^{\text{12}}}{\text{\hspace{0.33em}ms}}^{\text{-1}}\\ {\text{= 3.48307 × 10}}^{\text{5}}{\text{ms}}^{\text{-1}}\\ {\text{= 348 kms}}^{\text{-1}}\end{array}$

Q.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Ans.

$\begin{array}{l}{\text{Given, V}}_{\text{ο}}\text{= 1.5 V}\\ \text{Thus maximum kinetic energy}\\ {\text{K.E}}_{\text{max}}{\text{= eV}}_{\text{0}}\\ {\text{= 1.6 × 10}}^{\text{-19}}\text{C × 1.5 V}\\ {\text{K.E}}_{\text{max}}{\text{= 2.4 × 10}}^{\text{-19}}\text{J\hspace{0.17em}}\end{array}$

Q.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \text{Given,}\\ \text{λ = 632.8\hspace{0.17em}\hspace{0.33em}nm}\\ {\text{= 632.8 × 10}}^{\text{-9}}\text{\hspace{0.17em}m}\\ \text{Power, P = 9.42 \hspace{0.17em}mW}\\ {\text{= 9.42 × 10}}^{\text{-3}}\text{\hspace{0.17em}W}\\ \text{As energy of each photon is\hspace{0.33em}E =}\frac{\text{hc}}{\text{λ}}\\ \text{E =}\frac{\left({\text{6.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}\right)}{{\text{632.8 × 10}}^{\text{-9}}\text{\hspace{0.33em}m}}\\ {\text{= 3.14 × 10}}^{\text{-19}}\text{J}\\ \text{Momentum of photon}\\ \text{p =}\frac{\text{h}}{\text{λ}}\\ \text{=}\frac{{\text{6.626 × 10}}^{\text{-34}}\text{\hspace{0.33em}Js}}{{\text{632.8 × 10}}^{\text{-9}}\text{\hspace{0.33em}m}}\\ {\text{p = 1.05 ×10}}^{\text{-27}}{\text{kgms}}^{\text{-1}}\\ \left(\text{b}\right)\\ \text{If N is number of photons emitted per second, then power P transmitted in the beam equals N}\\ \text{times the energy per photon so that}\\ \text{N =}\frac{\text{P}}{\text{E}}\\ \text{=}\frac{{\text{9.42 × 10}}^{\text{-3}}\text{W}}{{\text{3.14 × 10}}^{\text{-19}}\text{J}}\\ {\text{= 3 × 10}}^{\text{16}}\text{\hspace{0.33em}photons\hspace{0.33em}per\hspace{0.33em}second\hspace{0.17em}}\\ \left(\text{c}\right)\\ {\text{As mass of hydrogen atom is equivalent to mass of proton which is 1.67 x 10}}^{\text{-27}}\\ \text{p = mv}\\ \text{v =}\frac{\text{p}}{\text{m}}\\ \text{Thus,\hspace{0.33em}v =}\frac{{\text{1.05×10}}^{\text{-27}}{\text{\hspace{0.33em}kgms}}^{\text{-1}}}{{\text{1.67×10}}^{\text{-27}}\text{\hspace{0.33em}kg}}\\ {\text{= 0.63 m\hspace{0.17em}s}}^{\text{-1}}\end{array}$

Q.5 The energy flux of sunlight reaching the surface of the earth is 1.388 x 10³ Wm^-2. How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Ans.

\begin{array}{l}\text{Given,}\\ \text{λ = 550 nm}\\ {\text{= 550 × 10}}^{\text{-9}}\text{nm}\\ \text{h = 6}{\text{.626 × 10}}^{\text{-34}}\text{Js}\\ {\text{c = 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}\\ \text{Energy}\text{flux,}\text{Φ = 1}{\text{.388 × 10}}^{\text{3}}{\text{Wm}}^{\text{-2}}\\ \text{Hence,}\text{power}\text{of}\text{sunlight}\text{per}\text{square}\text{metre,}\text{P = 1}{\text{.388 × 10}}^{\text{3}}{\text{Wm}}^{\text{-2}}\\ \text{Thus number of photons incident per square metre of earth}\\ \text{per second will be}\\ \text{N =}\frac{\text{P}}{\text{E}}\\ \text{=}\frac{\text{Pλ}}{\text{hc}}\frac{\text{1}{\text{.388 × 10}}^{\text{3}}{\text{Wm}}^{\text{-2}}{\text{× 550 × 10}}^{\text{-9}}\text{m}}{\text{6}{\text{.626 ×10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}\\ \text{= 3}{\text{.85 × 10}}^{\text{21}}{\text{photons/m}}^{\text{2}}\text{s}\end{array}

Q.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10^-15 Vs. Calculate the value of Planck’s constant.

Ans.

$\begin{array}{l}\text{The slope of the cut-off voltage versus frequency of incident}\\ \text{light is}\\ \frac{{\text{V}}_{\text{ο}}}{\mathrm{\nu }}{\text{= 4.12 × 10}}^{\text{-15}}\text{Vs}\\ \text{As per photoelectric equation}\\ {\text{eV}}_{\text{ο}}\text{= h}\mathrm{\nu }\text{–}\mathrm{\varphi }\\ \text{thus, the slope of the graph in this case is}\\ \frac{{\text{V}}_{\text{ο}}}{\mathrm{\nu }}\text{=}\frac{\text{h}}{\text{e}}\\ \text{which implies that}\\ \frac{\text{h}}{\text{e}}{\text{= 4.12 × 10}}^{\text{-15}}\text{Vs}\\ \text{Thus,}\\ {\text{h = 4.12 × 10}}^{\text{-15}}\text{Vs\hspace{0.33em}× e}\\ {\text{= 4.12 × 10}}^{\text{-15}}{\text{Vs\hspace{0.33em}× 1.6 × 10}}^{\text{-19}}\text{\hspace{0.33em}C}\\ {\text{Hence, h = 6.592 × 10}}^{\text{-34}}\text{Js \hspace{0.17em}}\end{array}$

Q.7 A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with the sodium light?
(b) At what rate are the photons delivered to the sphere?

Ans.

$\begin{array}{l}\text{(a)}\\ \text{Power of sodium lamp, P = 100 W}\\ \text{λ = 589 nm}\\ {\text{= 589 × 10}}^{\text{-9}}\text{m}\\ \text{Energy of photon}\\ \text{E =}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{{\text{6.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{{\text{589 × 10}}^{\text{-9}}\text{\hspace{0.33em}m}}\\ {\text{Thus, E = 3.38 × 10}}^{\text{-19}}\text{J}\\ \text{= 2.11\hspace{0.17em}eV\hspace{0.17em}}\\ \text{(b)}\\ \text{Rate at which the photons are delivered is equal to\hspace{0.33em}number}\\ \text{of photons\hspace{0.33em}delivered per second\hspace{0.17em}.}\\ \text{Thus,\hspace{0.33em}N =}\frac{\text{P}}{\text{E}}\text{\hspace{0.33em}\hspace{0.33em}=}\frac{{\text{100 kgms}}^{\text{-1}}}{{\text{3.38 × 10}}^{\text{-19}}\text{\hspace{0.33em}J}}\\ {\text{N = 2.96 × 10}}^{\text{20}}{\text{\hspace{0.17em}photons s}}^{\text{-1}}\end{array}$

Q.8 The threshold frequency for a certain metal is 3.3 x 10¹⁴ Hz. If light of frequency 8.2 x 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Ans.

$\begin{array}{l}\text{Given that threshold frequency}\\ {\nu }_o\text{= 3}{\text{.3 × 10}}^{\text{14}}\text{Hz}\\ \text{Incident frequency}\\ \nu \text{= 8}{\text{.2 × 10}}^{\text{14}}\text{Hz}\\ \text{h = 6}{\text{.626 × 10}}^{\text{-34}}\text{Js}\\ \text{e = 1}{\text{.6 × 10}}^{\text{-19}}\text{C}\\ \text{Using photoelectric equation,}\\ {\text{eV}}_o\text{= h}\nu \text{– h}{\nu }_o\\ {\text{V}}_{\text{o}}\text{=}\frac{\text{h}}{\text{e}\left(\nu \text{–}{\nu }_o\right)}\\ {\text{V}}_{\text{o}}\text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\text{1}{\text{.6 × 10}}^{\text{-19}}\text{C ×}\left(\text{8}{\text{.2 × 10}}^{\text{14}}\text{Hz – 3}{\text{.3 × 10}}^{\text{14}}\text{Hz}\right)}\\ {\text{Thus, V}}_o\text{= 2}\text{.03 V}\end{array}$

Q.9 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Ans.

$\begin{array}{l}\text{Work function,}\\ {\mathrm{\varphi }}_{{}_{\text{ο}}}\text{\hspace{0.33em}= 4.2 eV}\\ {\text{= 4.2 × 1.6 × 10}}^{\text{-19}}\text{\hspace{0.17em}J}\\ {\text{= 6.72 × 10}}^{\text{-19}}\text{\hspace{0.17em}J}\\ \text{As per definition of work function}\\ {\mathrm{\varphi }}_{\text{ο}}\text{= h}{\mathrm{\nu }}_{\text{ο}}\\ \text{Thus,}\\ {\mathrm{\nu }}_{\text{ο}}\text{=}\frac{\text{W}}{\text{h}}\\ \text{=}\frac{{\text{6.72 × 10}}^{\text{-19}}\text{\hspace{0.33em}J}}{{\text{6.626 × 10}}^{\text{-34}}\text{\hspace{0.33em}Js}}\\ \text{Hence,}\\ {\mathrm{\nu }}_{\text{ο}}{\text{= 1.01 × 10}}^{\text{15}}\text{\hspace{0.17em}Hz}\\ {\text{As\hspace{0.33em}λ}}_{{}_{\text{ο}}}\text{=}\frac{\text{c}}{{\mathrm{\nu }}_{\text{ο}}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\mathrm{ms}}^{-1}}{{\text{1.01 × 10}}^{\text{15}}\text{\hspace{0.33em}Hz}}\\ {\text{Thus\hspace{0.33em}λ}}_{\text{ο}}{\text{= 2.97 × 10}}^{\text{-7}}\text{m}\\ {\text{= 297 × 10}}^{\text{-9}}\text{m}\\ \text{= 297 nm}\\ {\text{Since threshold wavelength\hspace{0.33em}λ}}_{\text{ο}}\text{\hspace{0.33em}<\hspace{0.33em}incident wavelength λ,}\\ \text{which implies that,}\\ \text{Threshold frequency \hspace{0.33em}}{\mathrm{\nu }}_{\text{ο}}\text{\hspace{0.33em}>\hspace{0.33em}}\mathrm{\nu }\text{,\hspace{0.33em}therefore the photoelectric}\\ \text{emission for this radiation cannot take place.}\end{array}$

Q.10 Light of frequency 7.21 x 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 10⁵ ms^-1 are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Ans.

$\begin{array}{l}\text{As per photoelectric equation}\\ \frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}{}_{\text{max}}\text{= h}\mathrm{\nu }\text{– h}{\mathrm{\nu }}_{\text{0}}\\ \text{h}{\mathrm{\nu }}_{\text{0}}\text{= h}\mathrm{\nu }\text{\hspace{0.33em}-\hspace{0.33em}}\frac{\text{1}}{\text{2}}{{\text{mv}}^{\text{2}}}_{\text{max}}\\ \text{Thus}\\ {\mathrm{\nu }}_{\text{0}}\text{=}\frac{\left(\text{h}\mathrm{\nu }\text{\hspace{0.33em}-\hspace{0.33em}}\frac{\text{1}}{\text{2}}{{\text{mv}}^{\text{2}}}_{\text{max}}\right)}{\text{h}}\\ \text{=}\frac{\left({\text{7.21 × 10}}^{\text{14}}\text{J –}\frac{\text{1}}{\text{2}}{\text{× 9.1 × 10}}^{\text{-31}}\text{kg ×}{\left({\text{6 × 10}}^{\text{5}}{\mathrm{ms}}^{-1}\right)}^{\text{2}}\right)}{{\text{6.626 × 10}}^{\text{-34}}\text{\hspace{0.33em}Js}}\\ {\text{= 7.21 × 10}}^{\text{14}}{\text{Hz – 2.47 × 10}}^{\text{14}}\text{\hspace{0.33em}Hz}\\ {\text{= 4.74 × 10}}^{\text{14}}\text{Hz\hspace{0.17em}}\end{array}$

Q.11 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made.

Ans.

$\begin{array}{l}\text{Given,}\\ {\text{λ = 488 × 10}}^{\text{-9}}\text{\hspace{0.33em}m}\\ \text{stopping\hspace{0.33em}potential\hspace{0.33em}of\hspace{0.33em}photoelectron\hspace{0.33em}=\hspace{0.33em}0.38\hspace{0.33em}eV}\\ \mathrm{\nu }\text{=}\frac{\text{c}}{\text{λ}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{{\text{488 × 10}}^{\text{-9}}\text{\hspace{0.33em}m}}\\ \text{Thus,\hspace{0.33em}}\mathrm{\nu }{\text{= 6.15 × 10}}^{\text{14}}\text{Hz}\\ \text{Using photoelectric equation}\\ {\text{eV}}_{\text{0}}\text{= h}\mathrm{\nu }\text{– W}\\ \text{Where W is work function of the material, we get}\\ \text{W = h}\mathrm{\nu }{\text{– eV}}_{\text{0}}\\ \text{W =}\left({\text{6.626 × 10}}^{\text{-34}}{\text{Js × 6.15 × 10}}^{\text{14}}\text{\hspace{0.33em}Hz}\right)\text{–}\left({\text{1.6 × 10}}^{\text{-19}}\text{C × 0.38 eV}\right)\\ {\text{W = 3.467 × 10}}^{\text{-19}}\text{J}\\ \text{Thus W in eV will be\hspace{0.33em}}\frac{{\text{3.467 × 10}}^{\text{-19}}\text{\hspace{0.33em}J}}{{\text{1.6 × 10}}^{\text{-19}}\text{\hspace{0.33em}C}}\text{= 2.167 eV}\end{array}$

Q.12 Calculate the
(a) momentum, and
(b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Ans.

\begin{array}{l}\text{(a) Given,}\\ \text{V = 56 V, thus for an electron of mass “m”, charge “e” on being accelerated by potential of 56 V,}\\ \text{the momentum “p” will be}\\ \text{p =}\sqrt{\text{2 meV}}\\ \text{p =}\sqrt{\text{2 × 9}{\text{.1 × 10}}^{\text{-31}}\text{kg × 1}{\text{.6 × 10}}^{\text{-19}}\text{C × 56 V}}\\ \text{p =}\sqrt{\text{1}\text{.630}{\text{.72 × 10}}^{\text{-50}}}{\text{kgms}}^{\text{-1}}\\ \text{p = 4}{\text{.038 × 10}}^{\text{-24}}{\text{kgms}}^{\text{-1}}\\ \text{(b)}\\ \text{As per de – Broglie’s relation wavelength of an electron will be}\\ \text{λ =}\frac{\text{12}\text{.27}}{\sqrt{\text{V}}}\text{Å}\\ \text{Thus,}\text{λ =}\frac{\text{12}\text{.27}}{\sqrt{\text{56 V}}}\text{Å}\\ \text{λ = 1}\text{.64 Å}\\ \text{= 0}\text{.164}\text{nm}\end{array}

Q.13 What is the
(a) momentum
(b) speed and
(c) de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Ans.

\begin{array}{l}\text{(a)}\\ \text{Given K}\text{.E = 120 eV = 120 ×}\text{1}\text{.6}\text{×}{\text{10}}^{\text{-19}}\\ \text{=}\text{1}{\text{.92 × 10}}^{\text{-17}}\text{J}\\ \text{thus for an electron of mass “m”, charge “e”, and the momentum “p” will}\\ \text{p =}\sqrt{\text{2}\text{m}\text{E}}\\ \text{=}\sqrt{\text{2 × 9}{\text{.1 × 10}}^{\text{-31}}\text{kg × 1}{\text{.92 × 10}}^{\text{-17}}\text{J}}\\ \text{=}\sqrt{\text{34}{\text{.94 × 10}}^{\text{-48}}}\text{kg}{\text{ms}}^{\text{-1}}\\ \text{= 5}{\text{.91 × 10}}^{\text{-24}}{\text{kgms}}^{\text{-1}}\\ \text{(b)}\\ \text{The speed of an electron of mass “m” and momentum “p” will be}\\ \text{v =}\frac{\text{p}}{\text{m}}\\ \text{Thus,}\text{v =}\frac{\text{5}{\text{.91 × 10}}^{\text{-24}}{\text{kgms}}^{\text{-1}}}{\text{9}{\text{.1 × 10}}^{\text{-31}}\text{kg}}\\ \text{= 6}{\text{.5 × 10}}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{(c)}\\ \text{de-Broglie wavelength “λ” of an electron of momentum “p”}\\ \text{will be, λ =}\frac{\text{h}}{\text{p}}\\ \text{λ =}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\text{5}{\text{.91 × 10}}^{\text{-24}}{\text{kgms}}^{\text{-1}}}\\ \text{= 1}{\text{.121 × 10}}^{\text{-10}}\text{m}\\ \text{= 1}\text{.121 Å = 0}\text{.1121}\text{nm}\end{array}

Q.14 The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron and
(b) a neutron, would have the same de-Broglie wavelength.

Ans.

\begin{array}{l}\text{(a)}\\ \text{For an electron,}\\ \text{λ =}\frac{\text{h}}{\sqrt{\text{2}{\text{m}}_{\text{e}}\text{E}}}\\ \text{squaring both sides}\\ {\text{λ}}^{\text{2}}\text{=}\frac{{\text{h}}^{\text{2}}}{\text{2}{\text{m}}_{\text{e}}\text{E}}\\ \text{E =}\frac{{\text{h}}^{\text{2}}}{\text{2}{\text{m}}_{\text{e}}{\text{λ}}^{\text{2}}}\\ \text{E =}\frac{{\left(\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}\right)}^{\text{2}}}{\text{2 × 9}{\text{.1 × 10}}^{\text{-31}}\text{kg ×}{\left({\text{589 × 10}}^{\text{-9}}\text{m}\right)}^{\text{2}}}\\ \text{= 6}{\text{.95 × 10}}^{\text{-25}}\text{J}\\ \text{(b)}\\ \text{For a neutron,}\\ \text{E =}\frac{{\text{h}}^{\text{2}}}{{\text{2m}}_{\text{n}}{\text{λ}}^{\text{2}}}\\ \text{=}\frac{{\left(\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}\right)}^{\text{2}}}{\text{2 × 1}{\text{.67 × 10}}^{\text{-27}}\text{kg ×}{\left({\text{589 × 10}}^{\text{-9}}\text{m}\right)}^{\text{2}}}\\ \text{E = 3}{\text{.79 × 10}}^{\text{-28}}\text{J}\end{array}

Q.15 What is the de-Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 kms^-1,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 ms^-1, and
(c) a dust particle of mass 1.0 x 10^-9 kg drifting with a speed of 2.2 ms^-1?

Ans.

\begin{array}{l}\begin{array}{l}\begin{array}{l}\text{(a)}\\ \text{Given mass of the bullet, m = 0}\text{.040 kg}\end{array}\hfill \\ \text{Velocity, v = 1}{\text{.0 kms}}^{\text{-1}}{\text{= 1000 ms}}^{\text{-1}}\hfill \end{array}\\ \text{Thus, wavelength}\\ \text{λ =}\frac{\text{h}}{\text{mv}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\text{0}{\text{.040 kg × 1000 ms}}^{\text{-1}}}\\ \text{= 1}{\text{.66 × 10}}^{\text{-35}}\text{m}\\ \text{(b)}\\ \text{For a ball of mass, m = 0}\text{.060 kg}\\ \text{Velocity, v = 1}{\text{.0 ms}}^{\text{-1}}\\ \text{λ =}\frac{\text{h}}{\text{mv}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\text{0}\text{.060 kg × 1}\text{.0}{\text{ms}}^{\text{-1}}}\\ \text{λ = 1}{\text{.10 × 10}}^{\text{-32}}\text{m}\\ \text{(c)}\\ \text{For a dust particle of mass, m = 1}{\text{.0 x 10}}^{\text{-9}}\text{kg}\\ \text{Velocity, v = 2}{\text{.2 ms}}^{\text{-1}}\\ \text{λ =}\frac{\text{h}}{\text{mv}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\text{1}{\text{.0 × 10}}^{\text{-19}}\text{kg ×2}{\text{.2 ms}}^{\text{-1}}}\\ \text{λ = 3}{\text{.01 × 10}}^{\text{-25}}\text{m}\end{array}

Q.16 An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta
(b) the energy of the photon and
(c) the kinetic energy of electron.

Ans.

\begin{array}{l}\text{(a)}\\ \text{Momentum of an electron and photon,with same wavelength,}\\ \text{will be same as}\\ \text{p =}\frac{\text{h}}{\text{λ}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{{\text{1 × 10}}^{\text{-9}}\text{m}}\\ \text{Thus, p = 6}{\text{.626 × 10}}^{\text{-2}}{}^{\text{5}}{\text{kgms}}^{\text{-1}}\\ \text{(b)}\\ \text{Energy of photon,}\\ \text{E =}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{1 × 10}}^{\text{-9}}\text{m}}\\ \text{Thus, E}\text{= 1}{\text{.99 × 10}}^{\text{-16}}\text{J}\\ \text{Energy of photon in eV}\\ \text{=}\frac{\text{1}{\text{.99 × 10}}^{\text{-16}}}{\text{1}{\text{.6 × 10}}^{\text{-16}}}\text{keV}\\ \text{E = 1}\text{.24 keV}\\ \text{(c)}\\ \text{Kinetic energy of electron}\\ \text{=}\frac{{\text{p}}^{\text{2}}}{\text{2m}}\\ \text{=}\frac{{\left(\text{6}{\text{.626 × 10}}^{\text{-25}}{\text{kgms}}^{\text{-1}}\right)}^{\text{2}}}{\text{2 × 9}{\text{.1 × 10}}^{\text{-31}}\text{kg}}\\ \text{K}\text{.}\text{E = 2}{\text{.412 × 10}}^{\text{-19}}\text{J}\\ \text{=}\frac{\text{2}{\text{.412 × 10}}^{\text{-19}}}{\text{1}{\text{.6 × 10}}^{\text{-19}}}\text{eV}\\ \text{= 1}\text{.51 eV}\end{array}

Q.17 (a) For what kinetic energy of a neutron will the associated de-Broglie wavelength be 1.40 x 10^-10 m?
(b) Also find the de-Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of

\frac{\text{3}}{\text{2}}

kT at 300 K.

Ans.

\begin{array}{l}\text{(a)}\\ \text{λ = 1}{\text{.40 × 10}}^{\text{-10}}\text{m}\\ \text{h = 6}{\text{.63 × 10}}^{\text{-34}}\text{J}\text{s}\\ \text{m = 1}{\text{.67 × 10}}^{\text{-27}}\text{kg}\\ \text{Thus, λ =}\frac{\text{h}}{\sqrt{\text{2mE}}}\\ \text{E =}\frac{{\text{h}}^{\text{2}}}{{\text{2mλ}}^{\text{2}}}\\ \text{E =}\frac{{\left(\text{6}{\text{.63×10}}^{\text{-34}}\text{Js}\right)}^{\text{2}}}{\text{2 × 1}{\text{.67 × 10}}^{\text{-27}}\text{kg ×}{\left(\text{1}{\text{.4×10}}^{\text{-10}}\text{m}\right)}^{\text{2}}}\\ \text{= 6}{\text{.7 × 10}}^{\text{-21}}\text{J}\\ \text{Energy in eV will be}\\ \text{E =}\frac{\text{6}{\text{.7 × 10}}^{\text{-21}}\text{J}}{\text{1}{\text{.6 × 10}}^{\text{-19}}\text{eV}}\\ \text{= 4}{\text{.19 × 10}}^{\text{-2}}\text{eV}\\ \text{(b)}\\ \text{As average kinetic energy}\\ \text{E =}\frac{\text{3}}{\text{2}}\text{kT}\\ \text{wavelength,}\text{λ =}\frac{\text{h}}{\sqrt{\text{2}\text{m}\text{E}}}\\ \text{=}\frac{\text{h}}{\sqrt{\text{3}\text{m}\text{k}\text{T}}}\\ \text{where,}\text{k = 1}{\text{.38 × 10}}^{\text{-23}}\text{J}{\text{K}}^{\text{-1}}\text{and temperature T = 300 K}\\ \text{Thus,}\text{λ =}\frac{\left(\text{6}{\text{.63 × 10}}^{\text{-34}}\text{Js}\right)}{\sqrt{\text{3 × 1}{\text{.67 × 10}}^{\text{-27}}\text{kg × 1}{\text{.38 × 10}}^{\text{-23}}\text{J}{\text{K}}^{\text{-1}}\text{× 300 K}}}\\ \text{λ = 1}{\text{.456 × 10}}^{\text{-10}}\text{m}\end{array}

Q.18 Show that the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its quantum (photon).

Ans.

$\begin{array}{l}{\text{The de-Broglie wavelength of photon is\hspace{0.33em}λ}}_{\text{1}}\text{=}\frac{\text{h}}{\text{p}}\\ \text{As for a photon \hspace{0.33em}p =}\frac{\text{h}\mathrm{\nu }}{\text{c}}\\ {\text{Thus,\hspace{0.33em}λ}}_{\text{1}}\text{=}\frac{\text{hc}}{\left(\frac{\text{h}\mathrm{\nu }}{\text{c}}\right)}\\ \text{=}\frac{\text{c}}{\mathrm{\nu }}\\ \text{= λ}\\ \text{= wavelength of electromagnetic radiation\hspace{0.17em}.}\end{array}$

Q.19 What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules a this temperature. (Atomic mass of nitrogen = 14.0076 u).

Ans.

\begin{array}{l}\text{As for nitrogen, rms velocity,}\text{v =}\frac{\sqrt{\text{3kT}}}{\text{m}}\\ \text{Thus, de-Broglie wavelength of nitrogen molecule will be}\\ \text{λ =}\frac{\text{h}}{\text{m}\text{v}}\\ \text{=}\frac{\text{h}}{\sqrt{\text{3}\text{m}\text{k}\text{T}}}\\ \text{For Nitrogen molecule,}\\ \text{Mass, m = 2 × 14}\text{.0076 u}\\ \text{= 28}\text{.01521}\text{u}\\ \text{= 28}\text{.01521 × 1}{\text{.66 × 10}}^{\text{-27}}\text{kg}\\ \text{Hence,}\text{λ =}\frac{\text{6}{\text{.63 × 10}}^{\text{-34}}Js}{\sqrt{\text{3 × 28}\text{.0152 × 1}{\text{.66 × 10}}^{\text{-27}}\text{kg × 1}{\text{.38 × 10}}^{\text{-23}}{\text{JK}}^{-1}\text{× 300 K}}}\\ \text{λ = 0}{\text{.275 × 10}}^{\text{-10}}\text{m}\\ \text{= 0}\text{.0275 nm}\end{array}

Q.20 (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., Its e/m is given to be 1.76 × 10¹¹ Ckg^-1.
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Ans.

$\begin{array}{l}\text{As given,}\\ \text{V = 500 V;}\\ \frac{\text{e}}{\text{m}}\text{= 1.76}\times {\text{10}}^{\text{11}}{\text{\hspace{0.17em}C kg}}^{\text{-1}}\\ \text{Thus, using\hspace{0.33em}eV =}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{\hspace{0.33em}or\hspace{0.33em}v =}\sqrt{\frac{\text{2eV}}{\text{m}}}\\ \text{v =}\sqrt{\text{2}\times \text{1.76}\times {\text{10}}^{\text{11}}{\text{C kg}}^{\text{-1}}\times \text{500 V}}\\ \text{= 1.33}\times {\text{10}}^{\text{7}}{\text{ms}}^{\text{-1}}\\ \left(\text{b}\right)\text{Using the same formula for V = 10 MV = 10}\times {\text{10}}^{\text{6}}\text{V}\\ \text{v}=\sqrt{\text{2}\times \text{1.76}\times {\text{10}}^{\text{11}}{\text{C kg}}^{\text{-1}}\times \text{1}0\times \text{1}{0}^{\text{6}}\text{V}}{}^{}\\ =\sqrt{\text{3}.\text{52}\times \text{\hspace{0.33em}1}{0}^{\text{18}}}{{\text{ms}}^{\text{-1}}}^{}\\ \text{So , v}=\text{1}.\text{88\hspace{0.33em}}\times \text{\hspace{0.33em}1}{0}^{\text{9}}{\text{ms}}^{\text{-1}}\\ \text{The formula employed in part}\left(\text{a}\right)\text{is not valid. The expression}\\ \text{for energy \hspace{0.33em}}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{\hspace{0.33em}can only be used in the non-relativistic limit.}\end{array}$

$$\begin{gathered} \mathrm{For}\mathrm{very}\mathrm{high}\mathrm{speed}\mathrm{are}\mathrm{concerned},\mathrm{the}\mathrm{relative}\mathrm{domains}\mathrm{comes} \\ \mathrm{into}\mathrm{consideration}.\mathrm{For}\mathrm{this}\mathrm{we}\mathrm{use}\mathrm{relativistic}\mathrm{formula}\mathrm{as}\mathrm{given}: \\ \begin{array}{l}\text{Here mass,\hspace{0.33em} m\hspace{0.33em}=\hspace{0.33em}}\frac{{\text{m}}_{\text{0}}}{\sqrt{\text{1-}\frac{{\text{v}}^{\text{2}}}{{\text{c}}^{\text{2}}}}}\text{\hspace{0.33em} is to be considered}\\ \text{So that, r =\hspace{0.33em}}\frac{{\text{m}}_{\text{0}}\text{v}}{\text{eB}\sqrt{\text{1 -\hspace{0.33em}}\frac{{\text{v}}^{\text{2}}}{{\text{c}}^{\text{2}}}}}\end{array} \end{gathered}$$

Q.21 (a) A monoenergetic electron beam with electron speed of 5.20 × 10⁶ ms^-1 is subject to a magnetic field of 1.30 × 10^-4 T normal to beam velocity. What is the radius of circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹ C kg^-1.
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

Ans.

\begin{array}{l}\left(\text{a}\right)\\ \text{As given}\\ \text{v = 5}\text{.20}\times {\text{10}}^{\text{6}}{\text{ms}}^{\text{-1}}\text{; B = 1}\text{.30}\times {\text{10}}^{\text{-4}}\text{T,}\\ \frac{\text{e}}{\text{m}}\text{= 1}\text{.76}\times {\text{10}}^{\text{11}}{\text{C kg}}^{\text{-1}}\\ \text{As the condition for a moving electron to trace a circle under the effect of magnetic field is,}\\ \text{q}\text{v}\text{B =}\frac{{\text{mv}}^{\text{2}}}{\text{r}}\\ \text{or r =}\frac{\text{m}\text{v}}{\text{q}\text{B}}\\ \text{=}\frac{\text{m}\text{v}}{\text{e}\text{B}}\left(\text{q = e}\right)\\ \text{Hence,}\\ \text{r =}\frac{\text{5}\text{.20}\times {\text{10}}^{\text{6}}m{s}^{-1}}{\text{1}\text{.76}\times {\text{10}}^{\text{11}}{\text{Ckg}}^{-1}\times \text{1}\text{.30}\times {\text{10}}^{\text{-4}}\text{T}}\\ \text{= 2}\text{.27}\times {\text{10}}^{\text{-1}}\text{m}\\ \text{= 22}\text{.7 cm}\\ \left(\text{b}\right)\\ \text{Energy, E = 20 MeV}\\ \text{= 20}\times {\text{10}}^{\text{6}}\text{eV}\times \text{1}{\text{.6 x 10}}^{\text{-19}}\text{J}\\ \text{The energy of the electron is given by:}\\ \text{E =}\frac{1}{2}{\text{mv}}^{\text{2}}\\ \text{v =}\sqrt{\frac{\text{2}\times \text{20}\times {\text{10}}^{\text{6}}\text{eV}\times \text{1}\text{.6}\times {\text{10}}^{\text{-19}}C}{\text{9}\times {\text{10}}^{\text{-31}}kg}}\\ \text{= 2}\text{.67}\times {\text{10}}^{\text{9}}{\text{ms}}^{\text{-1}}\\ \text{Which}\text{is}\text{greater}\text{than}\text{the}\text{velocity}\text{of}\text{light}\text{.}\end{array}

Q.22 An electron gun with its anode at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (10^-2 mm of Hg). A magnetic field of 2.83 x 10^-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method) Determine e/m from the data.

Ans.

$\begin{array}{l}\text{As given, r = 12.0 cm = 0.12 m}\\ \text{As kinetic energy gained by an electron under the effect of anode potential is\hspace{0.33em}\hspace{0.17em}}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{= eV}\\ {\text{Which implies that\hspace{0.33em}mv}}^{\text{2}}\text{= 2\hspace{0.17em}eV\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}.\dots \dots \dots \text{(1)}\\ \text{Also,\hspace{0.33em}}\frac{{\text{mv}}^{\text{2}}}{\text{r}}\text{=\hspace{0.33em} evB}\\ \text{Thus,\hspace{0.33em}}\frac{\text{2eV}}{\text{r}}\text{= evB}\\ \text{v =}\frac{\text{2\hspace{0.17em}V}}{\text{r\hspace{0.17em}B}}\\ \text{v =}\frac{\text{2}\times \text{100 V}}{\text{0.12 m}\times \text{2.83}\times {\text{10}}^{\text{-4}}\text{T}}\\ \text{v = 5.89}\times {\text{10}}^{\text{6}}{\text{\hspace{0.17em}ms}}^{\text{-1}}\\ \text{From eqn …}..\left(\text{1}\right)\\ \frac{\text{e}}{\text{m}}\text{=}\frac{{\text{v}}^{\text{2}}}{\text{2V}}\\ \text{= \hspace{0.17em}}\frac{{\left(\text{5.89}\times {\text{10}}^{\text{6}}{\mathrm{ms}}^{-1}\right)}^{\text{2}}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\times \text{100\hspace{0.33em}V}}\\ \text{= 1.73}\times {\text{10}}^{\text{11}}{\text{\hspace{0.17em}Ckg}}^{\text{-1}}\end{array}$

Q.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at

0.\text{45}\AA

. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube.

Ans.

$\begin{array}{l}\text{(a)}\\ \text{As given}\\ {\text{λ}}_{\text{min}}\text{= 0.45 Å}\\ {\text{= 0.45 × 10}}^{\text{-10}}\text{\hspace{0.17em}m}\\ \text{Thus maximum energy of a photon will be}\\ {\text{E}}_{\text{max}}\text{=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{{\text{6.626 × 10}}^{\text{-34}}{\text{\hspace{0.33em}Js × 3 × 10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{{\text{0.45 × 10}}^{\text{-10}}\text{\hspace{0.33em}m}}\\ {\text{= 4.42 × 10}}^{\text{-15}}\text{J}\\ \text{=}\frac{{\text{4.42 × 10}}^{\text{-15}}}{{\text{1.6 × 10}}^{\text{-16}}}\text{\hspace{0.17em} ke\hspace{0.17em}V}\\ \text{= 27.61 keV}\\ \text{(b)}\\ \text{In\hspace{0.33em}X-ray\hspace{0.33em}tube\hspace{0.33em}for\hspace{0.33em}getting\hspace{0.33em}X-\hspace{0.33em}ray\hspace{0.33em}photons\hspace{0.33em}of\hspace{0.33em}27.6\hspace{0.33em}keV,\hspace{0.33em}}\\ \text{the\hspace{0.33em}kinetic\hspace{0.33em}energy\hspace{0.33em}of\hspace{0.33em}atleast\hspace{0.33em}27.6\hspace{0.33em}keV\hspace{0.33em}is\hspace{0.33em}required.\hspace{0.33em}So,\hspace{0.33em}accelerating}\\ \text{voltage\hspace{0.33em}of\hspace{0.33em}30\hspace{0.33em}keV\hspace{0.33em}is\hspace{0.33em}required\hspace{0.33em}to\hspace{0.33em}applied\hspace{0.33em}across\hspace{0.33em}the\hspace{0.33em}X-ray\hspace{0.33em}tube\hspace{0.33em}to}\\ \text{get\hspace{0.33em}X-ray\hspace{0.33em}photons.}\\ \end{array}$

Q.24 In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ -rays of equal energy. What is the wavelength associated with each γ-ray?
(1 BeV = 10⁹ eV)

Ans.

$\begin{array}{l}\text{As given energy carried by the pair of\hspace{0.33em}γ-rays = 10.2 BeV}\\ \text{Hence, energy of each\hspace{0.33em}γ-ray is\hspace{0.33em}E =}\frac{\text{hc}}{\text{λ}}\\ \text{λ =}\frac{\text{hc}}{\text{E}}\\ \text{=}{\frac{{\text{6.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{{\text{8.16 × 10}}^{\text{-10}}\text{\hspace{0.33em}J}}}^{}\\ {\text{λ = 2.43×10}}^{\text{-16}}\text{\hspace{0.17em}m\hspace{0.17em}}\\ \text{E =}\frac{\text{10.2 BeV}}{\text{2}}{\text{= 5.1 BeV = 5.1×10}}^{\text{9}}\text{\hspace{0.17em}eV}\\ {\text{= 5.1 × 10}}^{\text{9}}{\text{eV × 1.6×10}}^{\text{-19}}\text{J}\\ {\text{= 8.16 ×10}}^{\text{-10}}\text{J \hspace{0.17em}}\end{array}$

Q.25 Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^-10 Wm^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 × 10¹⁴ Hz.

Ans.

$\begin{array}{l}\text{(a)}\\ \text{Number of photons emitted per second,}\\ \text{n =}\frac{\text{Power of transmitter}}{\text{Energy of each photon}}\\ \text{As, Power of transmitter is}\\ \text{P = 10 k\hspace{0.17em}W}\\ {\text{= 10}}^{\text{4}}\text{\hspace{0.17em}W and λ = 500 m}\\ \text{Using\hspace{0.17em},}\\ \text{E =}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{{\text{(6.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}\text{)}}{\text{500\hspace{0.33em}m}}\\ {\text{= 3.98 × 10}}^{\text{-28}}\text{J}\\ \text{Thus, \hspace{0.33em}n =}\frac{\text{P}}{\text{E}}\\ \text{=}\frac{{\text{10}}^{\text{4}}\text{\hspace{0.33em}W\hspace{0.33em}}}{{\text{3.98 × 10}}^{\text{-28}}\text{\hspace{0.33em}J}}\\ {\text{= 2.52 × 10}}^{\text{31}}{\text{s}}^{\text{-1}}\\ \text{As the energy of a radio photon is exceedingly small and}\\ \text{the number of photons emitted per second in a radio beam}\\ \text{is enormously large.}\\ \text{(b)}\\ \begin{array}{l}\text{This is quite a small number, but still large enough to be counted by us}\\ \text{comparison of cases}\left(\text{a}\right)\text{and}\left(\text{b}\right)\text{tells us that our eye cannot count the number of photons individually.}\\ \end{array}\end{array}$

Q.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~ 10⁵ Wm^-2) red light of wavelength 6328 Å produced by a He-Ne laser?

Ans.

$\begin{array}{l}{\text{Here\hspace{0.17em}, λ}}_{\text{u}}\text{\hspace{0.33em}=\hspace{0.33em}2271 Å\hspace{0.33em}}\\ \text{=\hspace{0.33em}2271}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}m}\\ {\text{V}}_{\text{0}}\text{\hspace{0.33em}= 1.3\hspace{0.17em}V}\\ \text{Using the photoelectric equation we get,}\\ {\text{eV}}_{\text{0}}\text{= hν – W or,\hspace{0.33em}}\\ {\text{W = hν – eV}}_{\text{0}}\\ \text{W =}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{6.63}\times {\text{10}}^{\text{-34}}\text{Js}\times \text{3}\times {\text{10}}^{\text{-8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{\text{2271}\times {\text{10}}^{\text{-10}}\text{\hspace{0.33em}m}}\text{– 1.6}\times {\text{\hspace{0.33em}10}}^{\text{-19}}\text{C}\times \text{1.3\hspace{0.33em}V}\\ \text{= 6.67}\times {\text{10}}^{\text{-19}}\text{J}\\ \text{=}\frac{\text{6.67}\times {\text{10}}^{\text{-19}}}{\text{1.6}\times {\text{10}}^{\text{-19}}}\text{\hspace{0.17em}\hspace{0.33em}eV\hspace{0.33em}}\\ \text{= \hspace{0.17em}4.2 eV}\\ {\text{Let\hspace{0.33em}ν}}_{\text{0}}\text{\hspace{0.33em}be\hspace{0.33em}the\hspace{0.33em}threshold\hspace{0.33em}frequeny\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}metal\hspace{0.17em}.}\\ {\text{W = hν}}_{\text{0}}\\ {\text{ν}}_{\text{0}}\text{=}\frac{\text{W}}{\text{h}}\\ \text{=}\frac{\text{6.67}\times {\text{10}}^{\text{-19}}\text{\hspace{0.33em}J}}{\text{6.63}\times {\text{10}}^{\text{-34}}\text{\hspace{0.33em}Js}}\\ \text{= 1.0}\times {\text{\hspace{0.33em}10}}^{\text{15}}\text{\hspace{0.33em}Hz\hspace{0.17em}}\\ {\text{Also, for red light, λ}}_{\text{r}}\text{= 6328 Å}\\ \text{= 6328}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}m}\\ \text{Thus,\hspace{0.33em}frequency of red light}\\ {\mathrm{\nu }}_{\text{r}}\text{=}\frac{\text{c}}{{\text{λ}}_{\text{r}}}\\ \text{=}\frac{\text{3}\times {\text{10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{\text{6328}\times {\text{10}}^{\text{-10}}\text{\hspace{0.33em}m}}\\ \text{= 4.74}\times {\text{10}}^{\text{14}}\text{Hz\hspace{0.17em}.}\\ \text{Since\hspace{0.17em},\hspace{0.33em}}{\mathrm{\nu }}_{\text{r}}\text{\hspace{0.33em}<}{\mathrm{\nu }}_{\text{0}}\text{\hspace{0.17em},\hspace{0.33em}therefore\hspace{0.17em},\hspace{0.33em}the\hspace{0.33em}photocell\hspace{0.33em}will\hspace{0.33em}not\hspace{0.33em}respond}\\ \text{to\hspace{0.33em}this\hspace{0.33em}red\hspace{0.33em}light\hspace{0.17em},\hspace{0.33em}produced\hspace{0.33em}by\hspace{0.33em}laser\hspace{0.17em}.}\end{array}$

Q.27 Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

Ans.

\begin{array}{l}\text{As given wavelength from neon lamp is}\\ \text{λ = 640}\text{.2 nm}\\ \text{= 640}{\text{.2 × 10}}^{\text{-9}}\text{m}\\ \text{Using photoelectric equation}\\ {\text{eV}}_{\text{0}}\text{= hν – W}\\ \text{=}\left[\left(\frac{\text{hc}}{\text{λ}}\right)\text{–}\text{W}\right]\\ \text{W =}\left[\left(\frac{\text{h}\text{c}}{\text{λ}}\right){\text{– eV}}_{\text{0}}\right]\\ \text{=}\left[\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{640}{\text{.2 × 10}}^{\text{-9}}\text{m}}\text{\_}\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{C × 0}\text{.54}\text{V}\right)\right]\\ \text{= 3}{\text{.10 × 10}}^{\text{-19}}\text{J – 8}{\text{.64 × 10}}^{\text{-20}}\text{J}\\ \text{=}\left[\text{3}\text{.10 × 0}\text{.864}\right]{\text{× 10}}^{\text{-19}}\text{J}\\ \text{= 2}{\text{.24 × 10}}^{\text{-19}}\text{J}\\ \text{Thus, Work function}\\ \text{W =}\frac{\text{2}{\text{.24 × 10}}^{\text{-19}}}{\text{1}{\text{.6 × 10}}^{\text{-19}}}\text{eV}\\ \text{= 1}\text{.40 eV}\\ \text{Also, for wavelength line of an iron source,}\\ \text{λ = 4}\text{.272 nm}\\ \text{= 427}{\text{.2 × 10}}^{\text{-9}}\text{m}\text{,}\\ \text{Using}\text{,}\\ \text{W =}\frac{\text{hc}}{\text{λ}}{\text{– eV}}_{\text{0}}\\ {\text{eV}}_{\text{0}}\text{=}\frac{\text{hc}}{\text{λ}}\text{– W}\\ {\text{V}}_{\text{0}}\text{=}\frac{\text{hc}}{\text{eλ}}\text{–}\frac{\text{W}}{\text{e}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{1}{\text{.6 × 10}}^{\text{-19}}\text{C × 427}{\text{.2 × 10}}^{\text{-9}}\text{m}}\text{–}\frac{\text{2}{\text{.24 × 10}}^{\text{-19}}\text{J}}{\text{1}{\text{.6 × 10}}^{\text{-19}}\text{C}}\\ \text{= 2}\text{.91}\text{V – 1}\text{.40 V}\\ \text{= 1}\text{.51 V}\\ \text{Hence}\text{,}\text{new}\text{stopping}\text{potential}\text{is}\text{1}\text{.51}\text{V}\text{.}\end{array}

Q.28

$\begin{array}{l}\text{A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission,}\\ \text{since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum.}\\ \text{In our experiment with rubidium photo-cell, the following lines from a mercury source were used:}\\ {\text{λ}}_{\text{1}}{\text{= 3650 Å\hspace{0.33em}, λ}}_{\text{2}}{\text{= 4047 Å , λ}}_{\text{3}}\text{= 4358 Å,}\\ {\text{λ}}_{\text{4}}{\text{= 5461\hspace{0.33em}Å\hspace{0.33em}, λ}}_{\text{5}}\text{= 6907 Å\hspace{0.17em}.}\\ \begin{array}{l}\text{The stopping voltages, respectively, were measured to be:}\\ {\text{V}}_{\text{01}}{\text{= 1.28 V, V}}_{\text{02}}{\text{= 0.95 V, V}}_{\text{03}}{\text{= 0.74 V, V}}_{\text{04}}{\text{= 0.16 V, V}}_{\text{05}}\text{= 0 V}\\ \left(\text{a}\right)\text{Determine the value of Planck’s constant h.}\\ \left(\text{b}\right)\text{Estimate the threshold frequency and work function for the material.}\end{array}\end{array}$

Ans.

\begin{array}{l}\text{(a)}\\ \text{From the Einstein photoelectric equation}\text{,}\\ {\text{eV}}_o\text{= h}\nu \text{– W}\\ \text{Which implies}\\ \frac{\text{hc}}{\text{λ}}{\text{– eV}}_o\text{= W}\\ \text{As,}{\nu }_{\text{1}}\text{=}\frac{\text{c}}{{\text{λ}}_{\text{1}}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{3650 × 10}}^{\text{-10}}\text{m}}\\ \text{= 8}{\text{.22 × 10}}^{\text{14}}\text{Hz}\\ \text{Similarly,}\\ {\nu }_{\text{2}}\text{=}\frac{\text{c}}{{\text{λ}}_{\text{2}}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{4047 × 10}}^{\text{-10}}\text{m}}\\ \text{= 7}{\text{.41 × 10}}^{\text{14}}\text{Hz}\\ {\nu }_{\text{3}}\text{=}\frac{\text{c}}{{\text{λ}}_{\text{3}}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{4358 × 10}}^{\text{-10}}\text{m}}\\ \text{= 6}{\text{.88 × 10}}^{\text{14}}\text{Hz}\\ {\nu }_{\text{4}}\text{=}\frac{\text{c}}{{\text{λ}}_{\text{4}}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{5461 × 10}}^{\text{-10}}\text{m}}\\ \text{= 5}{\text{.49 × 10}}^{\text{14}}\text{Hz}\\ {\nu }_{\text{5}}\text{=}\frac{\text{c}}{{\text{λ}}_{\text{5}}}\\ \text{=}\frac{{\text{3 ×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{6907 × 10}}^{\text{-10}}\text{m}}\\ \text{= 4}{\text{.34 × 10}}^{\text{14}}\text{Hz}\\ {\text{A graph can be plotted between V}}_{\text{0}}\text{and}\nu \text{which comes}\\ \text{out to be a straight line}\text{.}\\ \text{Then,}\frac{\text{h}}{\text{e}}\text{=}\frac{\text{ΔV}}{\text{Δ}\nu }\\ \text{=}\frac{\left(\text{1}\text{.28 V – 0}\text{.16 V}\right)}{\left(\text{8}\text{.22 – 5}\text{.50}\right){\text{× 10}}^{\text{14}}\text{Hz}}\\ \text{Thus,}\text{h =}\frac{\left(\text{1}{\text{.6 × 10}}^{\text{-19}}\text{C}\right)\left(\text{1}\text{.12}\text{V}\right)}{\left(\text{2}{\text{.726 × 10}}^{\text{14}}\text{Hz}\right)}\\ \text{= 6}{\text{.57 × 10}}^{\text{-34}}\text{Js}\\ \text{(b)}\\ \text{Now using ,}\frac{\text{hc}}{{\text{λ}}_{\text{2}}}\text{– eV}{}_{\text{02}}\text{= W}\\ \text{W =}\frac{\text{hc}}{{\text{λ}}_{\text{2}}}\text{–}{\text{eV}}_{\text{02}}\\ \text{Thus,}\text{W =}\frac{\text{6}{\text{.6 × 10}}^{\text{-34}}{\text{Js × 3× 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{4047 × 10}}^{\text{-10}}\text{m}}\text{– 1}{\text{.6 × 10}}^{\text{-19}}\text{C × 0}\text{.95 V}\\ \text{= 4}{\text{.89 × 10}}^{\text{-19}}\text{J – 1}{\text{.52 × 10}}^{\text{-19}}\text{J}\\ \text{= 3}{\text{.37 × 10}}^{\text{-19}}\text{J =}\frac{\text{3}{\text{.37 × 10}}^{\text{-19}}}{\text{1}{\text{.6 × 10}}^{\text{-19}}}\text{eV}\\ \text{= 2}\text{.11 eV}\\ \text{Also,}\\ {\nu }_{\text{0}}\text{=}\frac{\text{W}}{\text{h}}\\ \text{=}\frac{\text{3}{\text{.37 × 10}}^{\text{-19}}\text{J}}{\text{6}{\text{.6 × 10}}^{\text{-34}}\text{J}s}\\ \text{= 5}{\text{.11 × 10}}^{\text{14}}\text{Hz}\end{array}

Q.29 The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; MO: 4.17 eV;
Ni : 5.15 eV.
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Ans.

\begin{array}{l}\text{Here, wavelength of incident radiation is}\\ \text{λ = 3300 Å}\\ {\text{= 3300 ×10}}^{\text{-10}}\text{m}\\ \text{Distance of the laser from the photocell}\text{,}\text{r = 1}\text{m}\\ \text{New distance, r’= 50}\text{cm = 0}\text{.5}\text{m}\\ \text{Using the relation,}\\ \text{E = h}\nu \\ \text{=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}{\text{Js × 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{3300 × 10}}^{\text{-10}}\text{m}}\\ \text{= 6}{\text{.02 × 10}}^{\text{-19}}\text{J}\\ \text{=}\frac{\text{6}{\text{.02 × 10}}^{\text{-19}}}{\text{1}{\text{.6 × 10}}^{\text{-19}}}\text{eV}\\ \text{= 3}\text{.76 eV}\\ \begin{array}{l}\begin{array}{l}\text{Since, the energy E of the incident photon of light is greater}\\ \text{than the work functions of Na and K, thus photoelectric emission}\\ \text{will occur in these metals}\text{. As the work function of MO and Ni}\\ \text{is greater than the incident energy, thus emission will not occur}\\ \text{in these metals}\text{.}\end{array}\hfill \\ \begin{array}{l}\text{If laser is brought closer, the intensity}\text{of radiation increases}\\ \text{but this will not affect the result regarding Mo and Ni}\text{. However}\\ \text{photoelectric current from Na and K will increase in proportion}\\ \text{to intensity}\text{.}\end{array}\hfill \end{array}\end{array}

Q.30 Light of intensity 10^-5 Wm^-2 falls on a sodium photo-cell of surface area 2 cm² . Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

Ans.

$\begin{array}{l}{\text{Intensity\hspace{0.33em}of\hspace{0.17em}incident\hspace{0.33em}light,\hspace{0.33em}I = 10}}^{\text{-5}}{\text{\hspace{0.17em}Wm}}^{\text{-2}}\\ {\text{Surface\hspace{0.33em}area\hspace{0.33em}of\hspace{0.33em}sodium\hspace{0.33em}photocell,\hspace{0.33em}A = 2\hspace{0.17em}cm}}^{\text{2}}\\ {\text{= 2 × 10}}^{\text{-4}}{\text{\hspace{0.17em}m}}^{\text{2}}\\ \text{Work\hspace{0.33em}function\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}metal,\hspace{0.33em}W = 2\hspace{0.17em}eV}\\ {\text{= 2 × 1.6 × 10}}^{\text{-19}}\text{J}\\ \text{Number\hspace{0.33em}of\hspace{0.33em}layers\hspace{0.33em}of\hspace{0.33em}sodium\hspace{0.33em}that\hspace{0.33em}absorbs\hspace{0.33em}the}\\ \text{incident\hspace{0.33em}energy,\hspace{0.33em}n = 5}\\ \text{As\hspace{0.33em}we\hspace{0.33em}know\hspace{0.33em}that\hspace{0.33em}the\hspace{0.33em}effective\hspace{0.33em}atomic\hspace{0.17em}area\hspace{0.33em}of}\\ {\text{a\hspace{0.33em}sodium\hspace{0.33em}atom,\hspace{0.33em}A}}_{\text{e}}{\text{= 10}}^{\text{-20}}{\text{\hspace{0.17em}m}}^{\text{2}}\\ \text{Number\hspace{0.33em}of\hspace{0.33em}conduction\hspace{0.33em}electrons\hspace{0.33em}in\hspace{0.33em}five\hspace{0.33em}layers}\\ \text{=}\frac{\text{5 × area\hspace{0.33em}of\hspace{0.33em}one\hspace{0.33em}layer}}{\text{effective\hspace{0.33em}atomic\hspace{0.33em}area}}\\ \text{=}\frac{{\text{5 × 2 × 10}}^{\text{-4}}{\mathrm{m}}^2}{{\text{10}}^{\text{-20}}{\mathrm{m}}^2}{\text{= 10}}^{\text{17}}\\ \begin{array}{l}\text{The answer obtained implies that the time of emission of}\\ \text{electron is very large and is not in agreement with the}\\ {\text{observed time of emission which is approximately 10}}^{\text{-9}}\text{s.}\\ \text{Thus, wave-picture of radiation is in disagreement with}\\ \text{photo-electric emission.}\end{array}\end{array}$

Q.31 Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate, voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1Å , which is of the order of inter-atomic spacing in the lattice) (m_e = 9.1 × 10^-31 kg).

Ans.

$\begin{array}{l}{\text{λ = 1 A}}^{\text{0}}\\ {\text{= 10}}^{\text{-10}}\text{\hspace{0.17em}m}\\ {\text{For X ray, \hspace{0.33em}E}}_{\text{X}}\text{=}\frac{\text{hc}}{\text{λ}}\\ \text{=}\frac{{\text{6.626 × 10}}^{\text{-34}}{\text{\hspace{0.33em}Js × 3 × 10}}^{\text{8}}{\text{\hspace{0.33em}ms}}^{\text{-1}}}{{\text{10}}^{\text{-10}}\text{\hspace{0.33em}m}}\\ {\text{= 1.99 × 10}}^{\text{-15}}\text{J}\\ \text{=}\frac{{\text{1.99 × 10}}^{\text{-15}}}{{\text{1.6 × 10}}^{\text{-19}}}\text{eV}\\ {\text{= 1.24 × 10}}^{\text{4}}\text{eV}\\ \text{= 12.4 KeV}\\ \text{For the electron, as per de Broglie’s relation\hspace{0.17em},}\\ \text{λ =}\frac{\text{h}}{\text{P}}\\ \text{=}\frac{\text{h}}{\sqrt{{\text{2 m}}_{\text{e}}{\text{\hspace{0.17em}E}}_{\text{e}}}}\\ \text{Squaring both the sides,}\\ {\text{E}}_{\text{e}}\text{=}\frac{{\text{h}}^{\text{2}}}{{\text{2m}}_{\text{e}}{\text{λ}}^{\text{2}}}\\ \text{=}\frac{{\left({\text{6.626 × 10}}^{\text{-34}}\text{\hspace{0.33em}Js}\right)}^{\text{2}}}{{\text{2 × 9.11 × 10}}^{\text{-31}}\text{\hspace{0.33em}kg ×}{\left({\text{10}}^{\text{-10}}\text{\hspace{0.33em}m}\right)}^{\text{2}}}\\ {\text{E}}_{\text{e}}{\text{= 2.41 × 10}}^{\text{-17}}\text{J}\\ \text{=}\frac{{\text{2.41 × 10}}^{\text{-17}}}{{\text{1.6 × 10}}^{\text{-19}}}\text{eV}\\ \text{= 151 eV}\\ \text{Clearly, the energy of photon is much greater}\\ \text{than the energy of electron\hspace{0.17em}.}\end{array}$

Q.32 (a) Obtain the de-Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. ( m_n = 1.675 × 10^-27 kg)
(b ) Obtain the de-Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Ans.

\begin{array}{l}\text{(a) Kinetic energy,}\\ \text{E = 150 eV}\\ \text{= 150 × 1}{\text{.6 × 10}}^{\text{-19}}\text{J}\\ \text{= 2}{\text{.4 × 10}}^{\text{-17}}\text{J}\\ \text{Using ,}\\ \text{λ =}\frac{\text{h}}{\sqrt{{\text{2 m}}_{\text{n}}\text{E}}}\\ \text{λ =}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\sqrt{\text{2 × 1}{\text{.675 × 10}}^{\text{-27}}\text{kg × 2}{\text{.4 × 10}}^{\text{-17}}\text{J}}}\\ \text{= 2}{\text{.34 × 10}}^{\text{-12}}\text{m}\\ \text{The}\text{inter-atomic}\text{spacing ~}\text{1 Å}\text{i}\text{.e}\text{.,}{\text{10}}^{\text{-10}}\text{m}\text{is}\text{about}\\ \text{a}\text{hundered}\text{times}\text{greater than the wave length}\text{.}\text{Hence,}\text{a}\text{neutron}\text{beam}\text{of}\\ \text{energy}\text{150}\text{eV}\text{is}\text{not}\text{sutiable}\text{for}\text{diffraction}\text{experiments}\text{.}\\ {\text{(b)Here T = 27}}^{\text{o}}\text{C}\\ \text{= 300 K}\\ \text{λ =}\frac{\text{h}}{\sqrt{\text{3}{\text{m}}_{\text{n}}\text{KT}}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\sqrt{\text{3 × 1}{\text{.675 × 10}}^{\text{-27}}\text{kg × 1}{\text{.38 × 10}}^{\text{-23}}{\text{J mole}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{× 300 K}}}\\ \text{= 1}{\text{.45 × 10}}^{\text{-10}}\text{m}\\ \text{= 1}\text{.45}\text{Å}\end{array}

Q.33 An electron microscope uses electron accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc,) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Ans.

\begin{array}{l}\text{Accelerating voltage,}\\ \text{V = 50 kV}\\ {\text{= 50 × 10}}^{\text{3}}\text{V}\\ \text{Wavelength of electron}\\ {\text{λ}}_{\text{e}}\text{=}\frac{\text{h}}{\sqrt{\text{2 meV}}}\\ \text{=}\frac{\text{6}{\text{.626 × 10}}^{\text{-34}}\text{Js}}{\sqrt{\text{2 × 9}{\text{.1 × 10}}^{\text{-31}}\text{kg × 1}{\text{.6 × 10}}^{\text{-19}}{\text{C × 50 × 10}}^{\text{3}}\text{V}}}\\ \text{= 5}{\text{.49 × 10}}^{\text{-12}}\text{m}\\ \text{= 5}{\text{.49 × 10}}^{\text{-12}}{\text{× 10}}^{\text{10}}\text{Å}\\ \text{= 0}\text{.0549 Å}\\ \text{Also, for yellow light,}\\ {\text{λ}}_{\text{y}}\text{= 5900 Å}\\ {\text{= 5900 × 10}}^{\text{-10}}\text{m}\\ \text{Now Resolving power, µ}\alpha \frac{\text{1}}{\text{λ}}\\ \frac{\text{Resolving power of electron microscope}}{\text{Resolving power of optical microscope}}\\ \text{=}\frac{{\text{λ}}_{\text{y}}}{{\text{λ}}_{\text{e}}}\\ \text{=}\frac{{\text{5900 × 10}}^{\text{-10}}\text{}\text{m}}{\text{0}{\text{.0549 × 10}}^{\text{-10}}\text{m}}\\ \text{= 1}{\text{.07 × 10}}^{\text{5}}\\ \text{Thus,}\text{the}\text{resolving}\text{power}\text{of}\text{an}\text{electron}\text{microscope}\\ \text{is}\text{nearly}{\text{10}}^{\text{5}}\text{times}\text{that}\text{of}\text{an}\text{optical}\text{microscope}\text{.}\end{array}

Q.34 The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^-15 m or less. This structure was first probe in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV).

Ans.

$\begin{array}{l}\text{Rest mass energy}\\ {\text{m}}_{\text{o}}{\text{c}}^{\text{2}}\text{= 0.511 MeV}\\ {\text{= 0.511 × 1.6 × 10}}^{\text{-13}}\text{J}\\ {\text{= 8.18 × 10}}^{\text{-14}}\text{J}\\ \text{Using de-Broglie’s relation\hspace{0.17em},}\\ \text{λ =}\frac{\text{h}}{\text{p}}\\ \text{p =}\frac{\text{h}}{\text{λ}}\\ \text{=}\frac{{\text{6.626 × 10}}^{\text{-34}}\text{\hspace{0.33em}Js}}{{\text{10}}^{\text{-15}}\text{\hspace{0.33em}m}}\\ {\text{= 6.626 × 10}}^{\text{-19}}{\text{kgms}}^{\text{-1}}\\ \text{As total energy of the particle is}\\ \text{E =}\sqrt{{\text{p}}^{\text{2}}{\text{c}}^{\text{2}}\text{+}{\left({\text{m}}_{\text{o}}{\text{c}}^{\text{2}}\right)}^{\text{2}}}\\ \text{=}\sqrt{\left[{\left({\text{6.626 × 10}}^{\text{-19}}{\text{\hspace{0.33em}kgms}}^{\text{-1}}\right)}^{\text{2}}\text{×}{\left({\text{3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}\right]\text{+}{\left({\text{8.18 × 10}}^{\text{-14}}\text{\hspace{0.33em}J}\right)}^{\text{2}}}\\ {\text{= 1.99 × 10}}^{\text{-10}}\text{J}\\ \text{=}\frac{{\text{1.99 × 10}}^{\text{-10}}}{{\text{1.6 × 10}}^{\text{-19}}}\text{eV}\\ {\text{= 1.24 × 10}}^{\text{9}}\text{eV ,}\left({\text{10}}^{\text{9}}\text{eV\hspace{0.33em}=\hspace{0.33em}1 BeV}\right)\\ \text{= 1.24 BeV}\\ \text{Thus, the energy of the proton ejected out of the}\\ \text{linear accelerator is of the order of\hspace{0.33em}a\hspace{0.33em}few BeV.}\end{array}$

Q.35 Find the typical de-Broglie wavelength associated with a He atom in helium gas at room temperature ( 27^ºC ) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

Ans.

$\begin{array}{l}{\text{Given, T = 27}}^{\text{0}}\text{C}\\ {\text{= 27}}^{\text{0}}\text{C + 273}\\ \text{= 300 K}\\ \text{P = 1 atm}\\ {\text{= 1.01 × 10}}^{\text{5}}{\text{\hspace{0.17em}N\hspace{0.17em}m}}^{\text{-2}}\\ \text{Mass of helium atom,}\\ \text{m =}\frac{\text{4}}{\text{Avogadro’s number}}\\ \text{=}\frac{\text{4}}{{\text{6.02 × 10}}^{\text{23}}}\text{g}\\ {\text{= 6.64 × 10}}^{\text{-24}}\text{g}\\ \text{Or}\\ {\text{m = 6.64 × 10}}^{\text{-27}}\text{kg ‘}\\ \text{As, λ =}\frac{\mathrm{h}}{\sqrt{\text{\hspace{0.17em}3\hspace{0.17em}mkT}}}\\ =\frac{{\text{6.626 × 10}}^{\text{-34}}\text{Js}}{\sqrt{{\text{3 × 6.64 × 10}}^{\text{-27}}{\text{kg × 1.38 × 10}}^{\text{-23}}{\text{J mole}}^{-1}{\text{K}}^{-1}\text{× 300 K}}}\\ {\text{= 7.3 × 10}}^{\text{-11}}\text{m}\\ {\text{= 7.3 × 10}}^{\text{-11}}{\text{× 10}}^{\text{10}}\text{Å}\\ \text{= 0.73 Å}\\ \text{Also, the mean separation}\\ {\text{r}}_{\text{0}}\text{=}{\left[\frac{\text{V}}{\text{N}}\right]}^{\text{1/3}}\\ \text{=}{\left[\frac{\text{kT}}{\text{P}}\right]}^{\text{1/3}}\\ \text{=}{\left[\frac{{\text{1.38 × 10}}^{\text{-23}}{\text{J mole}}^{-1}{\text{K}}^{-1}\text{× 300 K}}{{\text{1.01 × 10}}^{\text{5}}\mathrm{Pa}}\right]}^{\text{1/3}}\\ \text{=}{\left[{\text{4.1 × 10}}^{\text{-26}}\right]}^{\text{1/3}}\text{m}\\ {\text{= 3.45 × 10}}^{\text{-9}}\text{m}\\ {\text{= 3.45 × 10}}^{\text{-9}}{\text{× 10}}^{\text{10}}\text{Å}\\ \text{= 3.45 Å}\\ \frac{\text{λ}}{{\text{r}}_{\mathrm{o}}}\text{=}\frac{{\text{0.73 × 10}}^{\text{-10}}\mathrm{m}}{{\text{3.4 × 10}}^{\text{-9\hspace{0.33em}}}\mathrm{m}}\\ \text{= 0.021\hspace{0.17em}}\\ {\text{Clearly,\hspace{0.33em} r}}_{\mathrm{o}}\text{\hspace{0.33em}>> λ\hspace{0.17em}}\end{array}$

Q.36 Compute the typical de-Broglie wavelength of an electron in a metal at 27^ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10^-10 m.

Ans.

$\begin{array}{l}\text{Given,}\\ {\text{T = \hspace{0.33em}27\hspace{0.17em}}}^{\text{o}}\text{C}\\ {\text{= 27\hspace{0.17em}}}^{\text{o}}\text{C + 273}\\ \text{= 300 K}\\ \text{For\hspace{0.33em}an\hspace{0.33em}electron\hspace{0.33em}in\hspace{0.33em}metal\hspace{0.33em}}\\ \text{momentum,\hspace{0.33em}p =}\sqrt{\text{3\hspace{0.17em}m\hspace{0.17em}k\hspace{0.17em}T}}\\ \text{Now,}\\ \text{λ =}\frac{\text{h}}{\sqrt{\text{3\hspace{0.17em}mkT}}}\\ \text{=}\frac{{\text{6.626 × 10}}^{\text{-34}}\text{\hspace{0.33em}Js}}{\sqrt{{\text{3 × 9.1 × 10}}^{\text{-31}}{\text{kg × 1.38 × 10}}^{\text{-23}}{\text{J mole}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{× 300 K}}}\\ {\text{= 6.23 × 10}}^{\text{-9}}\text{m}\\ \frac{\text{λ}}{{\text{r}}_{\text{o}}}\text{=}\frac{{\text{6.23 × 10}}^{\text{-9}}\text{\hspace{0.33em}m}}{{\text{2 × 10}}^{\text{-10}}\text{\hspace{0.33em}m}}\\ \text{= 31.2}\\ {\text{where, r}}_{\text{o}}\text{\hspace{0.33em}is\hspace{0.33em}the\hspace{0.33em}mean\hspace{0.33em}separation\hspace{0.33em}between\hspace{0.33em}two\hspace{0.33em}electrons\hspace{0.33em}in\hspace{0.33em}a\hspace{0.33em}metal.}\\ {\text{Clearly λ >> r}}_{\text{o}}\end{array}$

Q.37 Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3) e;(-1/3) e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m ? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:

$\text{E}=\text{h}\mathrm{v},\text{\hspace{0.17em}p}=\frac{\text{h}}{\mathrm{\lambda }}$

but while the value of λ is physically significant, the value of ν  (and therefore, the value of the phase speed

$\mathrm{\nu }\mathrm{\lambda }$

) has no physical significance. Why?

Ans.

$\begin{array}{l}\text{(a)}\\ \text{In case of the Millikan oil drop experiment, the charge on the electron is measured.}\\ \text{The electron revolves outside the nucleus and each has a charge “e”.}\\ \text{Thus, we do not observe the fractional charges.}\\ \left(\text{i.e.\hspace{0.33em}+}\frac{\text{2}}{\text{3}}\text{e or –}\frac{\text{1}}{\text{3}}\text{e}\right)\\ \text{(b)}\\ \text{The energy acquired by the electron}\\ \text{when accelerated through V volt is}\\ \frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{= eV}\\ \text{v =}\sqrt{\frac{\text{2e V}}{\text{m}}}\\ \text{Similarly, the force on the electron in an electric field is}\\ {\text{F}}_{\text{e}}\text{= ma}\\ \text{= eE}\\ \text{a =}\frac{\text{e E}}{\text{m}}\\ \text{And force on electron in a magnetic field}\\ {\text{f}}_{\text{m}}\text{= evB}\\ \text{=}\frac{{\text{mv}}^{\text{2}}}{\text{r}}\\ \text{or v =}\frac{\text{erB}}{\text{m}}\\ \text{Clearly, the dynamic of electrons is determined by\hspace{0.33em}}\frac{\text{e}}{\text{m}}\\ \text{in all the above relations instead of “e” alone\hspace{0.17em}.}\\ \text{(c)}\\ \text{\hspace{0.33em}At low pressure, the density of gas decreases and mean free path of gas molecules increases. Hence,}\\ \text{ions do not collide frequently and are able to reach the respective electrodes to constitute a current.}\\ \text{At ordinary pressure, molecules of gas keep on colliding with each other and the ions formed do not}\\ \text{have a chance to reach the respective electrodes to constitute a current because of their recombination.}\\ \text{(d)}\\ \text{Work function in fact is the energy required to knock out the electron from highest filled level of}\\ \text{conduction band, there are different energy levels which collectively form a continuous band of levels.}\\ \text{Therefore, different amounts of energy are required to bring the electrons out of the different levels.}\\ \text{Electrons once omitted have different kinetic energies according to the energy supplied to the emitter.}\\ \text{(e)}\\ \text{Since, frequency for a given matter wave remains constant for different layers of the matter but}\\ \text{wavelength changes so is more significant than. The phase speed is not significant as group speed is}\\ \text{given by}\\ \frac{\text{dυ}}{\text{d(}\frac{\text{1}}{\text{λ}}\text{)}}\text{=}\frac{\text{dE}}{\text{dp}}\\ \text{=}\frac{\text{d}\left(\frac{{\text{p}}^{\text{2}}}{\text{2m}}\right)}{\text{dp}}\\ \text{=}\frac{\text{p}}{\text{m}}\\ \text{is physically significant\hspace{0.17em}.}\end{array}$