# CBSE Important Questions Class 12 Physics Chapter 13

## Important Questions for CBSE Class 12 Physics Chapter 13 – Nuclei

These important questions for class 12 Physics chapter 13 will teach students about nuclei. After reviewing these CBSE revision notes, students will become confident about answering questions in this chapter.

The entire mass of any atom is concentrated in the nucleus. An atom’s tiny nucleus is encircled by light electrons that spin around it without affecting the area between the limits of the nucleus and the atoms. Investigate the subject in depth to learn more about the nucleus of atoms. Students would also be better prepared for their exams by studying these chapter 13 class 12 Physics important questions. After studying these Physics class 12 chapter 13 important questions, they could solve several CBSE sample papers.

These notes also come with important formulas and CBSE extra questions that can be used to assess your comprehension.

### Study Important Questions for Class 12 Physics Chapter 13 – Nuclei

1. Describe the rationale for using heavy water as a moderator in nuclear reactors.

Ans: If a fission-produced neutron collides with a nucleus of the same mass, they slow down. Normal water can be utilized as a moderator because it includes hydrogen atoms, which have a mass almost equal to that of neutrons. However, through reaction, it quickly absorbs neutrons:

n + p d +

Deuteron is used here. Heavy water is employed as a moderator to get around this problem because it has a very low neutron absorption cross-section.

1. What absorbent substance does a nuclear reactor employ to regulate the rate at which neutrons react?

1. Describe the distinctive features of nuclear force.

Ans: (i) The most potent force in nature is nuclear force.

(ii) They are saturated forces.

(iii) They are charge-independent.

1. The mass number of two nuclei is 1:3. How do their nuclear densities compare?

Ans: The ratio of nuclear densities will be 1:1 because nuclear density is independent of mass number.

1. The mass numbers of two nuclei are in the proportion 2:5. How do their nuclear densities compare?

Ans: Because nuclear density and mass number are independent, the ratio will always be 1:1.

1. Why is nuclear fusion impractical in a lab setting?

Ans: Since it is impossible to maintain a temperature of 107 K in a laboratory therefore nuclear fusion is not possible in a lab setting.

1. Complete the nuclear reactions below:

(a)  Be49 + H11 Li36 + ……

(b)  B510 + He24 N713 + …..

Ans: (a) Be49 + H11 Li36 + He24

(b) B510 + He24 N713+ n01

1. What is a nuclear reaction’s Q-value?

Ans: Q – value = (Mass of reactants – Mass of products)

1. Wavelengths of spectral lines obtained in the H spectrum are 9546 A0, 6463, A0 and 1216 A0. Which of these wavelengths is a member of the Lyman family?

Ans: 1216 A0 belongs to the Lyman series.

1. Write the empirical relation for the hydrogen atom’s Paschen series of lines.

Ans: 1 = R 1321n2

where n = 4,5,6,7,….

1. How much tritium will still be around in 25 years? Tritium has a half-life of 12.5 years.

Ans: Given,

t = 25 years, T = 12.5 years

NNo = (12)t/T

= (12)25/12.5

= (12)2

= 14

= 0.25

1. Determine an electron’s kinetic and potential energies in the hydrogen atom’s first orbit. (Given e = 1.6 X 10-19 C and r = 0.53 X 10-10 m).

Ans: Kinetic energy, K.E. = ke22r

= (1.6 X 10-19)2 X 9 X 1092 X 0.53 X 10-10

= 21.74 X 10-19 J

= 21.74 X 10-191.6 X 10-19

= 13.59 eV

Potential energy, P.E. = – ke2r = -2 K.E.

= -2 X 13.59

= -27.18 eV

Hence, K.E. is 13.59 eV, and P.E. is -27.18 eV.

1. Why is nuclear fusion in a laboratory impossible?

Ans: Nuclear fusion requires high temperatures, so it cannot be conducted in a laboratory. Therefore, a laboratory cannot produce it.

1. Calculate the equivalent energy in electron volts of 16 mg of mass.

Ans: We know, E =mc2

E = 16 X 10-6 kg X (3X108 m/s)2

E = 16 X 9 X 1010 Joules

E = 16 X 9 X  10101.6 X 10-19 eV

E = 9X1030 eV

1. Given the three isotopes of neon gas: Ne1020, Ne1021 and Ne1022 possess respective abundances of 90.51%, 0.27%, and 9.22%. The atomic masses of three isotopes are 19.99u, 20.99u, and 21.99u. Calculate the average atomic mass of neon.

Ans: Given,

Atomic mass of Ne1020, m1 = 19.99u

Abundance of Ne1020, 1 = 90.51%

Atomic mass of Ne10, 21m2 = 20.99u

Abundance of Ne1021, 2 = 0.27%

Atomic mass of Ne1022, m3 = 21.99u

Abundance of Ne1022, 3= 9.22%

Hence, the average atomic mass of neon is

m = m1 + m22m331+2+3

m = 19.99 X 90.51 + 20.99 X 0.27 + 21.99 X 9.2290.51 + 0.2 + 9.22

m = 20.1771u

1. From the given relation, R = R0A1/3, where R0 is constant and A is the mass number of the given nucleus, obtain the nearly constant nuclear matter density (i.e., independence from A).

Ans: We know,

R = R0A1/3

Where, R0 = constant

Nuclear matter density, = mass of nucleusvolume of nucleus

Consider m as the average mass of the nucleus.

Mass of the nucleus = mA

= mA43R3

= 3mA4(R0A1/3)3 = 3mA4R03A

= 3m4R03

Hence, the density of nuclear matter is unrelated to A and is constant.

1. (a) Draw the plot of the binding energy per nucleon (BE/A) vs mass number (A). Then, write down two key inferences that can be made about the nature of nuclear force.

(b) This graph describes how energy is released during nuclear fusion and fission.

(c) Summarize the fundamental nuclear mechanism of neutron decay. Why is it so challenging to detect neutrinos?

Ans:

Conclusion:

(i) The force is powerful enough to provide a few MeV of binding energy for each nucleon since it is attracting.

(ii) For mass numbers 2 to 20, there are sharply defined peaks corresponding to He24, C612, O816, etc. The peak indicates that their nuclei are relatively mass stable than the other nuclei in their neighbourhood.

(b) (i) We discover an increase in total binding energy and, consequently, a release of energy when we move from the region of heavy nuclei to the middle region. This suggests that nuclear fission—splitting a heavy nucleus into two about equal pieces—can liberate energy.

(ii) Similarly, we discover an increase in total binding energy and, thus, a release of energy when we move from lighter to heavier nuclei. This suggests nuclear fusion, the method by which two lighter nuclei fuse to form a heavier nucleus that can release energy.

(c) In beta decay, a neutron splits into a proton, electron, and neutrino since it is challenging to detect neutrinos because they are massless and chargeless.

1. He23 + He23 He24 + H11 + H11 + 12.86 MeV

(a) Even though it is conserved on both sides of the reaction, energy is nonetheless released. How? Explain.

(b) Plot the potential energy as a function of the distance between two nucleons. Then, identify the areas with potential energy (i) positive and (ii) negative.

Ans: (a) This difference in binding energy manifests as energy being released because the overall binding energies of the nuclei on the left and right sides of the reaction are different.

For separation (r) ≤ 0.8 fermi

Force is repulsive.

For r > 0.8, fermi force will be attractive.

1. (a) In reaction,  H12 + H12 He23 + n + 3.27  MeV.

Explain how the nucleon is conserved and energy is released.

(b) Also prove that nuclear density is independent of mass number.

Ans: (a) The law of nucleon number conservation is observed in all types of nuclear processes. However, it was discovered during the reaction that the mass of the finished product was somewhat lower than the total mass of the reactant components. Mass defect is the difference in mass between a nucleus and its parts. Therefore, energy is released according to the mass-energy relation E=(∆M)c2. The combined masses of the two deutrons in the reaction are greater than the combined masses of helium and neutron. Therefore, the mass defect’s energy equivalent is released.

(b) Nuclear density = Mass of nucleus Volume = mA43R3

where, (m= mass of each nucleon, R0= 1.2 X 10-15 m)

As, R = R0A1/3

Nuclear density= 3mA4R03A = 3m4R03

i.e., Independent of mass number A.

1. Answer the following questions, giving reasons:

(i) Why is it that nuclei with a mass number (A) between 30 and 170 are found to have a constant binding energy per nucleon?

(ii) Energy is released during the breakdown of a heavy nucleus with mass number  A = 240 into two nuclei, A = 120.

(iii) It is discovered that it is very challenging to experimentally detect neutrinos (or antineutrinos) in -decay.

Ans: (i) Nuclear weapons have a limited range. A sufficiently massive nucleus will allow some of a specific nucleon’s neighbours within the nuclear force’s range to affect it. The saturation property of the nuclear force is the characteristic that a specific nucleon only affects nucleons adjacent to it.

(ii) The parent nucleus’s binding energy per nucleon is lower than that of the two offspring nuclei. This procedure results in the release of this enhanced binding energy.

(iii) Neutrinos are massless, chargeless particles with little interaction with other particles. As a result, they may only go through a lot of material if they are seen.

1. Plot the potential energy of a pair of nucleons vs. the distance between them. Note the areas where the nuclear force is present.

(i) Attractive and

(ii) Repulsive.

Mention any two defining characteristics of nuclear forces.

Ans: The graph shows that at a distance of r0 = 1 FM, the attractive force between the two nucleons is at its strongest. Furthermore, the force is strongly attractive for separations greater than r0 and repulsive for separations less than r0.

The following are two characteristics of nuclear forces:

(i) The strongest relationship

(ii) Close-quarters combat,

(iii) Character with charge independence

1. Describe the radioactive decay law. Plot a graph for a given radioactive sample with a half-life of T1/2 that displays an undecayed nuclei’s number (N) as a function of time (t). Plot the amount of still healthy nuclei at (i) t = 3 T1/2 and (ii) t = 51/2.

Ans: The number of nuclei in the sample at any given instant is proportional to the number of nuclei undergoing decay per time unit. The graph depicts the number of unaltered nuclei as a function of time.

1. Plot the changing binding energy per nucleon against the mass number (A). This graphic describes how nuclear fusion and fission processes release energy.

Ans: 1. Nuclear fission:

Heavier nuclei are less stable because their binding energy per nucleon is lower than that of the medium. The B.E./nucleon shifts (increases) from roughly 7.6 MeV to 8.4 MeV when a heavier nucleus breaks into the lighter nuclei. Energy is released due to the product nuclei’s higher binding energy. This occurs during nuclear fission, the process that creates the atom bomb.

1. Nuclear fusion :

The binding energy per nucleon is low for light nuclei, making them less stable. As a result, energy is released when two light nuclei fuse to form a heavier nucleus due to the latter’s larger binding energy per nucleon.

1. Separate nuclear fusion from nuclear fission. Describe the two processes’ respective energy release mechanisms.

Ans: Nuclear fusion combines two light nuclei to create a heavier nucleus. In contrast, nuclear fission splits a heavy nucleus into smaller nuclei with the release of energy. Both involve converting some mass (or mass defect) into energy as per the equation: E= m X c2.

1. (a) Write down two characteristics of nuclear force.

(b) Plot the potential energy of a pair of nucleons against their distance from one another.

Ans: (a) Nuclear forces:

Nuclear forces are the potent forces of attraction that hold the nucleons (neutrons and protons) in the small nucleus of an atom together.

Important traits (properties):

(i) Charge does not affect nuclear forces (These act between pairs of neutrons, protons, proton and neutron).

(ii) The strongest forces in nature are nuclear forces.

(iii) Nuclear weapons have a relatively short range.

(iv) Non-central forces, such as nuclear forces.

(v)  Spin affects the nuclear forces.

(b) The graph depicts p.e. between two nucleons as a function of distance.

Important inferences from the graph:

(i) Compared to gravitational forces between masses or the Coulomb force operating between charges, the nuclear force is substantially greater. This is because the repulsive Coulomb force between protons inside the nucleus must yield to the nuclear binding force. This occurs only because the nuclear force is so much stronger than the coulomb force. Even the Coulomb force is substantially stronger than the gravitational force.

(ii) As soon as more than a few femtometers separate two nucleons, the nuclear force between them vanishes quickly. The constant binding energy per nucleon results from the saturation of forces in a medium- or large-sized nucleus.

(iii) Proton-neutron, proton-proton, and neutron-neutron nuclear forces are roughly equal. The electric charge has no bearing on the nuclear force.

1. How is mass transformed into energy (or vice versa) in a nuclear reaction if the quantity of protons and neutrons is conserved during each nuclear reaction? Explain.

Ans: In a nuclear process, the number of protons and neutrons is preserved, but the overall mass is not. Protons and neutrons are heavier in the free state than they are inside the nucleus as a whole. According to the equation, E = (∆m)c2, the “mass defect,” or lost mass (= ∆m), is transformed into energy.

1. Describe the rationale for using heavy water as a moderator in nuclear reactors.

Ans: If a fission-produced neutron collides with a nucleus of the same mass, they slow down. Normal water can be utilized as a moderator because it includes hydrogen atoms, which have a mass almost equal to that of neutrons. But through reaction, it quickly absorbs neutrons:

n + p d +

Here d is Deutron. Heavy water is utilized as a moderator to get around this problem because it has an external neutron absorption cross-section.

1. For every gram of carbon, biological stuff is shown to typically degrade roughly 15 times every minute. The modest amount of radioactive C614 coexists with the stable carbon isotopes C612, causing this activity. When an organism dies, it stops interacting with the environment, which keeps the above equilibrium activity going, and its activity declines. The specimen’s age can be roughly calculated using the observed activity and the known half-life of C614 (5730 years). This is the C614 dating principle that is applied in archaeology. Let’s say a sample from Mohenjo-Daro exhibits nine decays per minute per gram of carbon. Calculate the civilization’s age in the Indus Valley.

Ans: Given,

Rate of decay of the living carbon-containing matter, R= 15 decay/minute,

Let N represent the total number of radioactive atoms in a typical carbon-containing substance.

The half-life of C614, T1/2 = 5730 years.

The specimen from the Mohenjo-Daro site had a decay rate of R’ = 9 decays per minute.

Let N’ represent the sample’s radioactive atom count during the Mohenjo-Daro era.

Thus, the decay constant () and time (t) is related as: NN = RR = et

et = 915 = 35

t = loge 35 = -0.5108

t= 0.5108

Now, = 0.693T1/2 = 0.6935730

t= 0.51080.693/5730 = 4223.5

Thus, the appropriate age of Indus Valley civilization is about 42223.5 years.

1. Graph the relationship between nucleon binding energy and mass number. Name two prominent characteristics of the curve. Also, define binding energy.

Ans: The nucleus’s binding energy is the total energy needed to break the nucleus down into its individual particles.

Key characteristics of the curve:

(i) The intermediate nuclei are the most stable because they have a high binding energy per nucleon. (For 30 <A > 63)

(ii) The binding energy per nucleon is minimal for both light and heavy nuclei. They are unstable nuclei.

1. Define the decay constant of given radioactive samples. Which of the given radiations – rays, – rays, -rays.

(a) are similar to X-rays?

(b) are they easily absorbed by matter?

Ans: (a) The radioactive decay constant is the reciprocal of the amount of time needed for the radioactive substance’s atom count to fall to 36.8% of its initial value ().

– rays.  It is similar to that X-rays.

(b) The penetration power of -rays is lesser than – rays & – rays. Hence, -rays can be easily absorbed by matter.

1. Why does beta decay continuously generate beta particles with increasing energy?

Ans: The transformation of a neutron into the nucleus into a proton, electron, and antineutrino causes the beta decay process. The electron and antineutrino share the energy available in beta decay in all possible ratios when they exit the nucleus; hence the beta ray energy spectrum is continuous.

1. Describe how radioactive nuclei can emit β-particles despite the absence of these particles in atomic nuclei. Therefore, clarify why a radioactive nuclide’s mass number does not change throughout decay.

Ans: A nucleus does not contain beta particles (or electrons) as such. The following equation describes how a neutron can occasionally decay into a proton, an electron, and an antineutrino in the case of a radioactive nuclide:

n01 H11 + e-10 + v + Q

Where the antineutrino particle’s mass and charge are both zero, the proton from the produced particles stays inside the nucleus, while the electron and antineutrino leave the nucleus. This electron is the one emitting a beta particle.

The mass number of the nuclide doesn’t change during -decay because, similar to the emission process, one proton is created in the nucleus at the expense of a neutron since both have the same mass number.

1. Consider a radioactive nucleus A that goes through the A, B, and C steps to decay to a stable nucleus C. Here, the radioactive intermediate nucleus B is present. Plot a graph illustrating the change in the number of A and B atoms over time, assuming that there are initially N0 of A atoms.

Ans:  When time = 0, NA = No and NB = 0. The number of atoms in B grows to a maximum and then decays to zero , whereas NA decreases exponentially with time (following the exponential decay law).

Thus, the graph is as

1. How does mass become energy (or vice versa) in a nuclear reaction if the number of neutrons and protons is conserved in each nuclear reaction? Explain.

Ans: We know that a nucleus’ binding energy contributes negatively to the mass of the nucleus (mass defect). This is because the total rest mass of neutrons + protons is the same on both sides of a nuclear reaction because the number of protons and neutrons in a reaction is conserved. However, the total binding energy of the nuclei on the left and right need not be equal.

The energy released/absorbed in a nuclear reaction represents the difference between these binding energies. We argue that the difference in the total mass of the nuclei on the two sides gets transformed into energy or vice versa because binding energy contributes to mass.

1. Why are protons and neutrons never in greater abundance in stable nuclei?

Ans: Because they are positively charged, protons electrically repel one another. When there are more than 10 protons in a nucleus, this repulsion becomes so strong that stability depends on an excess of neutrons, which only produce attractive forces.

1. Is the resultant atom in beta decay when a nucleus experiences alpha decay electrically neutral?

Ans: No, because the atomic number drops by 2 in alpha decay, the atom retains 2 extra orbital electrons. Therefore it bears a double negative charge. Conversely, the atom retains a net positive charge after experiencing beta decay.

1. What other particle has a mass that is comparable to that of the proton? Neutrino, Neutron, Positron, Proton.

Ans: Let’s say the parent nucleus was P, which had a binding energy of 7.6 Mev per nucleon, and decayed into the daughter nucleus Q, which had a binding energy of 8.5 MeV per nucleon.

According to the question,

P Q+Q

Given, BE/A of element P = 7.6 MeV

So,  BE of P = 7.6 X 240 MeV

BE/A of element Q = 8.5 MeV

BE of Q = 8.5 X 120 MeV

Now, the energy released in the process will be a difference of total binding energy and can be computed as = 2(BE of Q) – BE of P

= 85 X 120 X 2-7.6 X 240

= 2040 – 1824 MeV

= 216 MeV

1. Plutonium has a half-life of 24000 years before it decays. What is the portion of plutonium left after 72000 years of storage?

Ans: Below is a representation of the binding energy per nucleon curve.

(i) The average binding energy per nucleon is roughly 8 MeV.

(ii) The most stable nuclei are those with a mass number of A = 50 or below, with the highest binding energy per nucleon. Close to the top is the iron nucleus Fe56, which has a binding energy per nucleon of roughly 8.8 MeV. One of the most stable nuclides in existence.

(iii) The binding energy per nucleon is lower in extremely low or extremely high mass number nuclei. As a result, they are less stable since breaking apart the nucleus into its nucleons is simpler.

(iv) Low-mass number nuclei can undergo nuclear fusion, in which light nuclei are fused under specific circumstances to produce a final product with a higher binding energy per nucleon.

1. How is the size of a nucleus measured experimentally? Describe the relationship between the nucleus’s mass number and radius. Establish the independence of the nucleus’ density from its mass number.

Ans: In nuclear physics, nuclear fission is the process by which an atom’s nucleus divides into two or smaller nuclei as fission products, along with typically some by-product particles. The total binding energy of the product produced by dissociating these radioactive nuclides is lower than that of the reactants, which release energy during radioactive decay.

U92235 + n01 Ba56141 + Kr3692+ 3n01 + Q

Here, the energy released per fission of U92215 is 200.4 MeV.

The energy emitted per fission of  U92215 in this instance is 200.4 MeV. Light nuclei will fuse with a yield of energy if they are pressed together because the combined mass will be smaller than the sum of the masses of the individual nuclei. The nuclear particles will be more tightly bonded than in the lighter nuclei if the overall nuclear mass is less than iron at the apex of the binding energy curve. According to the Einstein relationship, this drop in mass is released as energy.

e.g., H11 + H11 He12 + e+ + +0.42 MeV

H12 + H12 He13 + n + 3.27 MeV

H12 + H12 H13 + H11 + 4.03 MeV

As per the question,

H1 2+ H He24 + nMeV

Δm = (2014102 + 3.016049) – (4.002603 + 1.008665)

= 0.018883. u

Energy released, Q = 0.01883 X 9315 MeVc2 = 17.589 MeV

1. The mass numbers of two nuclei are 1:8. How do their nuclear radii compare?

Ans: Using Rutherford’s scattering experiment, along with the closest approach and impact parameters, the size of the nucleus can be experimentally determined. The results can be summarised as follows.

The relation between the radius and mass number of the nucleus is R= R0 A1/3

(here, R0= 1.2 fmR = radius of the nucleus and A= mass number of nuclear density)

= Mass of nucleusVolume of nucleus = mA43 (R0 A1/3)3

Where, m = mass of each nucleon (proton and neutron)

Therefore, according to the formula above, the density of the nucleus is independent of the nucleus’ mass number and is the same for all atoms. It is roughly 1017 kg/m3, which is extremely high compared to the densities we regularly observe.

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Q1- 92U238 after undergoing successively 8 alpha decays and 6 beta decays give

opt-

a-82U206

b-82Pb206

c-80Th200

d-82U180

ans-An alpha particle has a mass number 4 and atomic number 2 and a beta particle has a mass number 0 and atomic number -1. So, due to 8 alpha decays, the mass number decreases by 32 and atomic number reduces by 16. Similarly, due to 6 beta decays, the atomic number increases by 6.

Hence, the resulting nucleus is whose atomic number is 82[92 – 16 + 6] and mass number is 206[238-32], i.e., 82Pb206.

## 1. What connection exists between a nucleus's radius and its mass number A?

The relationship is R (A)1/3.

## 2. The mass numbers of two nuclei are in proportion 2:5. How do their nuclear densities compare?

Because nuclear density and mass number are independent, the ratio will always be 1:1.

## 3. Why is gamma rays' penetrating strength so high?

Because they are neutral and have tremendous energy.

## 4. What is meant by a radioactive substance's activity?

The activity of radioactive material is measured by how quickly it breaks down.

## 5. State relationship between mean life (τ) of any radioactive element and its decay constant .

The relation is τ = 1 / λ.