The Nucleus Multiple Choice Questions And Answers

The Nucleus

Question 1. If the radius of the \({ }_{13}^{27} \mathrm{Al}\) nucleus is taken to be RAl then the radius of the \({ }_{52}^{125} \mathrm{Te}\) nucleus is nearly

1. \(\left(\frac{52}{13}\right)^{1 / 3} R_{\mathrm{Al}}\)
2. \(\frac{5}{3} R_{\mathrm{Al}}\)
3. \(\frac{3}{5} R_{\mathrm{Al}}\)
4. \(\left(\frac{13}{52}\right)^{1 / 3} R_{\mathrm{Al}}\)

Answer: 2. \(\frac{5}{3} R_{\mathrm{Al}}\)

For a nucleus of mass number A, the radius is given by R = \(R_0 A^{1 / 3}\)

∴ \(\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}}=\left(\frac{125}{27}\right)^{1 / 3}=\frac{5}{3}\)

⇒ \(R_{\mathrm{Te}}=\frac{5}{3} R_{\mathrm{Al}}\)

Question 2. The mass number of helium is 4 and that for sulphur is 32. The radius of the sulfur nucleus is larger than the helium nucleus by a multiple of

1. 4
2. 2√2
3. 2
4. 8

Answer: 3. 2

Given that A_He = 4 and A_s = 32

∴ \(\frac{R_{\mathrm{S}}}{R_{\mathrm{He}}}=\left(\frac{32}{4}\right)^{1 / 3}=(8)^{1 / 3}=2\)

Hence, R_s = 2R_He.

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Question 3. The mass density of a nucleus varies with the mass number as

1. A
2. A²
3. \(\frac{1}{A}\)
4. A₀

Answer: 4. A₀

The mass density of a nucleus is given by

⇒ \(\rho=\frac{\text { mass }}{\text { volume }}=\frac{A m}{\frac{4}{3} \pi R^3}=\frac{A m}{\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3}=\frac{3 m}{4 \pi R_0}\)

which is constant and independent of A.

Thus, p ∝ A⁰.

Question 4. What is the radius of the iodine atom? (Given that atomic number = 53 and mass number = 126.)

1. 2.5 x 10^-11 m
2. 2.5 x 10^-9 m
3. 7 x 10^-9 m
4. 7 x 10^-6 m

Answer: 1. 2.5 x 10^-11 m

The electronic configuration of iodine is 2, 8, 18, 18, 7. Hence the principal quantum number of the outermost electron is n = 5.

∴ the radius of the outermost orbit is

⇒ \(R_n=R_0\left(\frac{n^2}{Z}\right)=(0.53 Å)\left(\frac{5^2}{53}\right)=0.25 Å=2.5 \times 10^{-11} \mathrm{~m}\)

Question 5. The nuclei \({ }_6^{13} \mathrm{C} \text { and }{ }_7^{14} \mathrm{~N}\) can be described as

1. Isotopes
2. Isotones
3. Isobars
4. None

Answer: 2. Isotones

In \({ }_6^{13} \mathrm{C}, n_{\mathrm{p}}\) = 6 and nn = 7.

In \({ }_7^{14} \mathrm{~N}, n_{\mathrm{p}}\) = 7 and «n = 7.

Both have the same number of neutrons (= 7).

Hence, they are isotones of each other.

Question 6. When an a-particle of mass m moving at a velocity v bombards on a heavy nucleus of charge Ze, its distance of the closest approach from the nucleus depends on m as

1. m
2. \(\frac{1}{\sqrt{m}}\)
3. \(\frac{1}{m}\)
4. \(\frac{1}{m^2}\)

Answer: 3. \(\frac{1}{m}\)

For an a-particle, charge = 4e.

For a heavy nucleus, charge = Ze.

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By the principle of energy conservation, the kinetic energy of the a-particle gets converted into its PE at the distance of the closest approach (d). Thus,

⇒ \(\frac{1}{2} m v^2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Z e)(4 e)}{d}\)

⇒ \(d=\frac{2 Z e^2}{\pi \varepsilon_0 m v^2} \Rightarrow d \propto \frac{1}{m}\)

Question 7. The binding energy of a deuteron is 2.2 MeV and that of a \({ }_2^4 \mathrm{He}\) nucleus is 28 MeV. If two deuterons are fused to form one \({ }_2^4 \mathrm{He}\) nucleus then the energy released is

1. 23.6 MeV
2. 30.2 MeV
3. 25.8 MeV
4. 19.2 MeV

Answer: 1. 23.6 MeV

A fusion reaction releases energy. The reaction is \({ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{Q}\)

Given that BE of a deuteron \(\left({ }_1^2 \mathrm{H}\right)=2.2 \mathrm{MeV}\) and BE of a helium \(\left({ }_2^4 \mathrm{He}\right) \text { nucleus }=28 \mathrm{MeV}\).

∴ the energy released is

Q = (BE of the daughter nuclei)- (BE of the parent nuclei)

= 28 MeV- 2(2.2 MeV) = 23.6 MeV.

Question 8. The binding energy per nucleon of the \({ }_3^7 \mathrm{Li} \text { and }{ }_2^4 \mathrm{He}\) nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}\) the value of the energy Q released is

1. 19.6 MeV
2. 17.3 MeV
3. -2.4 MeV
4. 8.4 MeV

Answer: 2. 17.3 MeV

The given nuclear reaction is \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}\)

Now, BE of the two \({ }_2^4 \mathrm{He} \text { nuclei }\) = 2(7.06 MeV) A_He = 7.06 MeV x 8

= 56.48 MeV.

and BE of the \({ }_3^7 \mathrm{Li} \text { nucleus }\) = 5.60 MeV x 7

= 39.20 MeV.

∴ the energy released is Q = 56.48 Mev = 17.28 Mev = 17.3 Mev.

Question 9. A certain mass of hydrogen is changed into helium by the process of fusion. The mass defect in this fusion reaction is 0.02866 u. The energy liberated per atomic mass unit is nearly (given that 1 u = 931 MeV)

1. 2.67 MeV
2. 6.67 MeV
3. 26.7 MeV
4. 13.35 MeV

Answer: 2. 6.67 MeV

Given that mass defect = m = 0.02866 u.

The energy released = E = mc².

∴ energy released per amu of \({ }_2^4 \mathrm{He}\) is

⇒ \(\frac{E}{4}=\frac{0.02866 \times 931 \mathrm{MeV}}{4 c^2} \cdot c^2=6.67 \mathrm{MeV}\)

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Question 10. The mass of the \({ }_3^7 \mathrm{Li}\) nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of the \({ }_3^7 \mathrm{Li}\) nucleus is nearly

1. 46 MeV
2. 3.9 MeV
3. 5.6 MeV
4. 23 MeV

Answer: 3. 5.6 MeV

Mass defect = Δm = 0.042 u.

So, the binding energy is

BE = Amc² = (0.042)(931 MeV).

∴ BE per nucleon = \(\frac{B E}{A}=\frac{0.042 \times 931 \mathrm{MeV}}{7}=5.58 \mathrm{MeV}=5.6 \mathrm{MeV}\).

Question 11. A nucleus \({ }_Z^A \mathrm{X}\) has a mass represented by m(A, Z). If mp and ran denote the masses of a proton and a neutron respectively, the binding energy is given by

1. \(\mathrm{BE}=\left[m(A, Z)-Z m_{\mathrm{p}}-(A-Z) m_{\mathrm{n}}\right] c^2\)
2. \(\mathrm{BE}=\left[\mathrm{Z} m_{\mathrm{p}}+(A-\mathrm{Z}) m_{\mathrm{n}}-m(A, \mathrm{Z})\right] c^2\)
3. \(\mathrm{BE}=\left[\mathrm{Z} m_{\mathrm{p}}+A m_{\mathrm{n}}-m(A, \mathrm{Z})\right] c^2\)
4. \(\mathrm{BE}=m(A, Z)-Z m_{\mathrm{p}}-(A-Z) m_{\mathrm{n}}\)

Answer: 2. \(\mathrm{BE}=\left[\mathrm{Z} m_{\mathrm{p}}+(A-\mathrm{Z}) m_{\mathrm{n}}-m(A, \mathrm{Z})\right] c^2\)

For the nucleus \({ }_Z^A \mathrm{X}\), the number of protons = Z and the number of neutrons = A-Z.

∴ the total mass of its constituents = initial mass

= Mi = Zm_p + (A-Z)m_n,

where wp and mÿ are the masses of a proton and a neutron respectively. When all the neutrons and protons combine to form a nucleus of mass M, the loss of mass, AM appears as the binding energy of the nucleus.

Thus, BE = [Zm_p + (A-Z)m_n – m(A, Z)]c².

Question 12. In the reaction \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n}\), if the binding energies of \({ }_1^2 \mathrm{H},{ }_1^3 \mathrm{H} \text { and }{ }_2^4 \mathrm{He}\) are respectively a, b and c then the energy released in this reaction is

1. c + a – b
2. c – a – b
3. a + b – c
4. a + b + c

Answer: 2. c – a – b

The given reaction is \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 \mathrm{n}\)

In this nuclear fusion reaction, the reactants \(\left({ }_1^2 \mathrm{H} \text { and }{ }_1^3 \mathrm{H}\right)\) fuse together to form a more stable nucleus \(\left({ }_2^4 \mathrm{He}\right)\) having more binding energy with the release of an energy Q.

∴ energy released = (BE of the final product)- (BE of the reactants)

= c – (a + b) = c – a – b.

Question 13. If in a nuclear fusion process, the masses of the fusing nuclei are m₁ and m₂ and the mass of the resultant nucleus is m₃ then

1. m₃ = m₁ + m₂
2. m3 > m₁ + m₂
3. m₃ < m₁ + m₂
4. m₃ = m₁ – m₂

Answer: 3. m₃ < m₁ + m₂

In a nuclear fusion reaction, the final mass of the fused nucleus is less than the total mass of the fused nuclei. This loss of mass appears as the energy released.

∴ m₁+ m₂ > m₃ => m₃ < m₁ + m₂.

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Question 14. The nuclear reaction \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_{\mathrm{Z}+1}^A \mathrm{Y}+{ }_{-1}^0 \mathrm{e}+\overline{\mathrm{v}}\) represents a

1. Fission
2. Fusion
3. γ-decay
4. β-decay

Answer: 4. β-decay

In the given reaction, an electron \(\left({ }_{-1}^0 \mathrm{e}\right)\) and an antineutrino \((\bar{v})\) are emitted. Such a reaction represents a β-decay.

Question 15. A fission of nuclei is possible because of the binding energy per nucleon in them

1. Increases with the mass number for high mass numbers
2. Increases with the mass number for low mass numbers
3. Decreases with the mass number for high mass numbers
4. Decreases with the mass number for low mass numbers

Answer: 3. Decreases with the mass number for high mass numbers

From the curve showing the BE per nucleon versus the mass number (A), it is observed that the nuclei with higher values of A (like \({ }^{235} \mathrm{U}\)) have lower values of BE/A, and they are thus unstable and can get fissioned easily. Further, for such heavier nuclei, BE/A decreases with an increase in A.

Question 16. Let m_p denote the mass of a proton and m_n that of a neutron. A given nucleus of binding energy BE contains Z protons and N neutrons. The mass of the nucleus is given by

1. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{P}}-(\mathrm{BE}) c^2\)
2. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{P}}+(\mathrm{BE}) c^2\)
3. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{p}}-\frac{\mathrm{BE}}{c^2}\)
4. \(m(N, Z)=N m_{\mathrm{n}}+\mathrm{Z} m_{\mathrm{p}}+\frac{\mathrm{BE}}{c^2}\)

Answer: 3. \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{p}}-\frac{\mathrm{BE}}{c^2}\)

The binding energy of a nucleus with Z protons and N neutrons is given by

BE = Amc² – [Zm_p + Nm_n– m(N, Z)]c².

∴ \(\frac{\mathrm{BE}}{c^2}=Z m_{\mathrm{p}}+N m_{\mathrm{n}}-m(N, Z)\)

⇒ \(m(N, Z)=N m_{\mathrm{n}}+Z m_{\mathrm{p}}-\frac{\mathrm{BE}}{c^2}\)

Question 17. The mass of a proton is 1.0073 u and that of a neutron is 1.0087 u. The binding energy of \({ }_2^4 \mathrm{He}\) is (mass of helium nucleus being 4.0015 u)

1. 28.4 MeV
2. 0.061 keV
3. 0.0305 J
4. 0.0305 erg

Answer: 1. 28.4 MeV

Total mass of the constituents = 2(m_p + m_n)

= 2(1.0073 +1.0087) u = 4.032 u. J

Mass of the fHe nucleus = M = 4.0015 u.

∴ BE of the helium nucleus = (4.032 – 4.0015) x 931 MeV

= 0.0305 x 931 MeV = 28.4 MeV.

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Question 18. In the fission reaction \({ }_{92}^{236} \mathrm{U} \rightarrow{ }^{117} \mathrm{X}+{ }^{117} \mathrm{Y}+\mathrm{n}+\mathrm{n}\) the binding energy per nucleon of X and Y is 8.5 MeV, whereas that of \({ }_2^4 \mathrm{He}\) is 7.6 MeV. The total energy liberated will be about

1. 200 MeV
2. 2000 MeV
3. 2 MeV
4. 1 keV

Answer: 1. 200 MeV

The binding energy of the nuclei X and Y taken together is

BE_f = (8.5 + 8.5)(117 MeV) = 1989 MeV.

The binding energy of the \({ }^{236} \mathrm{U}\) nucleus is

BE_i = 236(7.6 MeV) = 1793.6 MeV.

the total energy liberated is

ΔE = BE_i – BE_f = (1989-1793.6) MeV

= 195.4 MeV ≈ 200 MeV.

Question 19. The average binding energy of a nucleon inside an atomic nucleus is about

1. 8 eV
2. 8 erg
3. 8 MeV
4. 8 J

Answer: 3. 8 MeV

The average binding energy of a nucleon (proton or neutron) is the average value of the BE per unit mass number (BE/A). The (BE/A)-versus-A curve has a broad maximum in the range from A = 30 to A =120. This corresponds to an average value of BE/A of about 8 MeV.

Question 20. The energy released in the fission of a single
\({ }_{92}^{235} \mathrm{U}\) nucleus is 200 MeV. The fission rate of a \({ }_{92}^{235} \mathrm{U}\)-filled nuclear reactor operating at a power level of 5 W is

1. 1.56 x l0¹⁰ s^-1
2. 1.56 X 10¹⁶ s^-1
3. 1.56 x l0¹¹ s^-1
4. 1.56 x l0¹⁷ s^-1

Answer: 3. 1.56 x l0¹¹ s^-1

Output power = (number of fissions of \({ }^{235} \mathrm{U}\) per unit time)(200 MeV)

⇒ 5 J s^-1 = N(200MeV)

∴ \(N=\frac{5 \mathrm{~J} \mathrm{~s}^{-1}}{\left(200 \times 10^6\right)\left(1.6 \times 10^{-19} \mathrm{~J}\right)}=\frac{5 \times 10^{11} \mathrm{~J} \mathrm{~s}^{-1}}{3.2 \mathrm{~J}}=1.56 \times 10^{11} \mathrm{~s}^{-1}\).

Question 21. The binding energy per nucleon is maximum in the case of

1. \({ }_2^4 \mathrm{He}\)
2. \({ }_{26}^{56} \mathrm{Fe}\)
3. \({ }_{56}^{141} \mathrm{Ba}\)
4. \({ }_{92}^{235} \mathrm{U}\)

Answer: 2. \({ }_{26}^{56} \mathrm{Fe}\)

In the (BE/A)-versus-A curve, the maximum occurs at (8.8 MeV) /A for \({ }_{26}^{56} \mathrm{Fe}\).

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Question 22. If the binding energy per nucleon in the \({ }_3^7 \mathrm{Li} \text { and }{ }_2^4 \mathrm{He}\) nuclei are respectively 5.60 MeV and 7.06 MeV, the energy of each proton in the reaction \({ }_3^7 \mathrm{Li}+\mathrm{p} \rightarrow 2{ }_2^4 \mathrm{He}\) is

1. 2.4 MeV
2. 17.3 MeV
3. 8.4 MeV
4. 19.6 MeV

Answer: 2. 17.3 MeV

The total BE in the \({ }_3^7 \mathrm{Li}\) nucleus is 7(5.60 MeV) = 39.20 MeV.

The total BE in the two \({ }_2^4 \mathrm{He}\) nuclei is 2(4 x 7.06 MeV) = 56.48 MeV.

KE of the proton = difference in BE

= 56.48 MeV – 39.20 MeV

= 17.28 MeV ≈ 17.3 MeV.

Question 23. The stable nucleus that has a radius half that of \({ }^{56} \mathrm{Fe}\) is

1. \({ }^7 \mathrm{Li}\)
2. \({ }^{21} \mathrm{Na}\)
3. \({ }^{16} \mathrm{~S}\)
4. \({ }^{40} \mathrm{~K}\)

Answer: 1. \({ }^7 \mathrm{Li}\)

The radius of a nucleus is R = R₀A ^1/3, where A is the mass number.

Here, \(\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3}\)

⇒ \(\frac{R}{R / 2}=\left(\frac{56}{A_2}\right)^{1 / 3} \Rightarrow 2^3=8=\frac{56}{A_2} \Rightarrow A_2=7\)

Thus, the required nucleus is \({ }^7 \mathrm{Li}\).

Question 24. If nip and mn are the masses of a proton and a neutron respectively, for an element of mass m having Z protons and N neutrons,

1. m > Zm_p + Nmn
2. m = Zm_p + Nm_n
3. m < Zm_p + Nm_n
4. m may be greater than, less than, or equal to Zm_p + Nm_n, depending on the nature of the element

Answer: 3. m < Zm_p + Nm_n

During the formation of a nucleus, the mass of the nucleus is less than that of its constituents, and the mass defect appears as energy. Thus,

m < Zm_p + Nm_n.

Question 25. In a nuclear fission reaction, energy is released because the

1. The total mass of the products is more than the mass of the nucleus
2. The total binding energy of the products formed due to the nuclear fission is more than that of the fissionable parent materials
3. The total binding energy of the products formed due to the nuclear fission is less than that of the fissionable parent materials
4. The combined mass of some particles is converted into energy

Answer: 2. Total binding energy of the products formed due to the nuclear fission is more than that of the fissionable parent materials

In a nuclear fission reaction, the products (or daughter nuclei) are more stable, because their binding energy per nucleon (BE/A) is increased. Thus, the total BE of the products formed is more than that of the reactants.

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Question 26. Which of the following are suitable for a fusion process?

1. Light nuclei
2. Heavy nuclei
3. Elements lying in the middle of the periodic table
4. Elements lying in the middle of the binding-energy curve

Answer: 1. Light nuclei

Lighter nuclei have smaller values of BE/A, hence they are unstable. At higher temperatures, light nuclei fuse so as to increase BE/A and become more stable. Hence, light nuclei are suitable for a fusion process.

Question 27. Solar energy is mainly caused due to the

1. Fission of the uranium nuclei present in the sun
2. Gravitational contraction
3. Fusion of protons during the synthesis of heavier elements
4. Burning of hydrogen in oxygen

Answer: 3. Fusion of protons during the synthesis of heavier elements

Solar energy is produced by a thermonuclear process in which lighter nuclei (or protons) fuse together to form heavier elements.

Question 28. In any fission process, the ratio of the total mass of the fission products to that of the parent nuclei

1. Is less than 1
2. Is greater than 1
3. Is equal to 1
4. Depends on the masses of the parent nuclei

Answer: 1. Is less than 1

In nuclear fission, a heavy nucleus breaks up into comparatively lighter nuclei with the release of energy. This energy liberated is due to the loss of mass. Hence, the total mass of the fission products is less than the total mass of the parent nuclei.

∴ \(\frac{\text { total mass of the fission products }}{\text { total mass of the parent nuclei }}<1\)

Question 29. A nucleus of uranium at rest decays into nuclei of thorium and helium. Then,

1. The helium nucleus has more kinetic energy than the thorium nucleus
2. The helium nucleus has less momentum than the thorium nucleus
3. The helium nucleus has more momentum than the thorium nucleus
4. The helium nucleus has less kinetic energy than the thorium nucleus

Answer: 1. The helium nucleus has more kinetic energy than the thorium nucleus

The given reaction is \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{90}^{232} \mathrm{Th}+{ }_2^4 \mathrm{He}\).

The initial momentum before the decay is zero.

By the principle of conservation of linear momentum, \(\left|p_{\mathrm{Th}}\right|=\left|p_{\mathrm{He}}\right|\), and they move in opposite directions. But KE = p2/2m. So,

⇒ \(\mathrm{KE}_{\mathrm{He}}=\frac{p^2}{2 m_{\mathrm{He}}} \text { and } \mathrm{KE}_{\mathrm{Th}}=\frac{p^2}{2 m_{\mathrm{Th}}}\)

∵ \(m_{\mathrm{He}}<m_{\mathrm{Th}}\),

∴ \(\mathrm{KE}_{\mathrm{He}}>\mathrm{KE}_{\mathrm{Th}}\).

Question 30. A nucleus \({ }_n^m \mathrm{X}\) emits an a-particle and two (1-particles. The resulting nucleus is

1. \({ }_n^{m-6} Z\)
2. \({ }_n^{m-4} \mathrm{X}\)
3. \({ }_{n-2}^{n-4} \mathrm{Y}\)
4. \({ }_{n-4}^{m-6} \mathrm{Z}\)

Answer: 2. \({ }_n^{m-4} \mathrm{X}\)

The given nuclear reaction is \({ }_n^m \mathrm{X} \stackrel{\alpha}{\longrightarrow}{ }_{n-2}^{m-4} \mathrm{Y} \stackrel{2 \beta}{\longrightarrow}{ }_n^{m-4} \mathrm{Z}\)

The elements Z and X have the same atomic number (= n), so they are identical.

Hence, the resulting nucleus is \({ }_n^{m-4} \mathrm{X}\).

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Question 31. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter nucleus is an

1. Isobar of the parent nucleus
2. The isomer of the parent nucleus
3. Isotone of the parent nucleus
4. Isotope of the parent nucleus

Answer: 4. Isotope of the parent nucleus

For simplicity, consider the decay with one a-particle and two p-particles. Thus, for a given nucleus \({ }_Z^A \mathrm{X}\) we have

⇒ \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_2^4 \alpha+{ }_{\mathrm{Z}-2}^{A-4} \mathrm{Y}\)

Further,

⇒ \({ }_{\mathrm{Z}-2}^{A-4} \mathrm{Y} \rightarrow{ }_2^{-1}{ }_{-1}^0 \beta+{ }_{\mathrm{Z}}^{A-4} \mathrm{Z}\)

Thus, \({ }_Z^A \mathrm{X} \text { and }{ }_Z^{A-4} \mathrm{Z}\) have the same atomic number (= Z) but different mass numbers (respectively A and A- 4). Hence, they are isotopes of each other.

Question 32. The nuclei of which of the following pairs of nuclei are isotones?

1. \({ }_{34}^{74} \mathrm{Se} \text { and }{ }_{31}^{71} \mathrm{Ga}\)
2. \({ }_{42}^{92} \mathrm{Mo} \text { and }{ }_{40}^{92} \mathrm{Zr}\)
3. \({ }_{38}^{84} \mathrm{Sr} \text { and }{ }_{38}^{86} \mathrm{Sr}\)
4. \({ }_{20}^{40} \mathrm{Ca} \text { and }{ }_{16}^{32} \mathrm{~S}\)

Answer: 1. \({ }_{34}^{74} \mathrm{Se} \text { and }{ }_{31}^{71} \mathrm{Ga}\)

Isotones are the nuclides that have the same number of neutrons, which means the same value of A-Z.

For, \({ }_{34}^{74} \mathrm{Se}\) A-Z = 74- 34 = 40.

For, \({ }_{31}^{71} \mathrm{Ga}\) A-Z = 71-31 = 40.

Hence, \({ }_{34}^{74} \mathrm{Se} \text { and }{ }_{31}^{71} \mathrm{Ga}\) are isotones of each other.

Question 33. An unstable nucleus of mass M emits a photon of frequency v, and the nucleus recoils. The recoil energy will be

1. zero
2. hv
3. \(\frac{h^2 v^2}{2 M c^2}\)
4. Mc²– hv

Answer: 3. \(\frac{h^2 v^2}{2 M c^2}\)

Since the nucleus is initially at rest, the total momentum of the photon and the nucleus is zero.

∴ momentum of the photon = \(\frac{h}{\lambda}=\frac{h v}{c}=p\)

∴ the recoil KE of the nucleus is

⇒ \(\frac{p^2}{2 M}=\frac{1}{2 M}\left(\frac{h v}{c}\right)^2=\frac{h^2 v^2}{2 M c^2}\).

Question 34. In the nuclear decay \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_{\mathrm{Z}+1}^A \mathrm{Y} \rightarrow{ }_{\mathrm{Z}-1}^{A-4} \mathrm{~B} \rightarrow{ }_{\mathrm{Z}-1}^{A-4} \mathrm{~B}\), the particles emitted, in the sequence, are

1. β, α, γ
2. γ, β, α
3. β, γ, α
4. α, β, γ

Answer: 1. β, α, γ

Consider the subsequent transitions in the same order below:

⇒ \({ }_{\mathrm{Z}}^A \mathrm{X} \rightarrow{ }_{\mathrm{Z}+1}^A \mathrm{Y}+{ }_{-1}^0 \beta\)  (β-emission),

⇒ \(\stackrel{\mathrm{Z}+1}{A} \mathrm{Y} \rightarrow{ }_{\mathrm{Z}-1}^{A-4} \mathrm{~B}+{ }_2^4 \mathrm{He}\)  (α-emission)

and \({ }_{Z-1}^{A-4} \mathrm{~B} \rightarrow{ }_{Z-1}^{A-4} \mathrm{~B}\)  (γ-emission).

Hence, the sequence is β, α, γ.

Question 35. The radius of a germanium (Ge) nuclide is measured to be twice the radius of a \({ }_4^9 \mathrm{~B}\) nuclide. The number of nucleons in the Ge nuclide is

1. 73
2. 74
3. 75
4. 72

Answer: 4. 72

The number of nucleons is the mass number (A) of the nucleus.

Now, R = R₀A^1/3

∴ \(\frac{R_{\mathrm{Ge}}}{R_{\mathrm{B}}}=\left(\frac{A}{9}\right)^{1 / 3}\)

⇒ \(\frac{2}{1}=\left(\frac{A}{9}\right)^{1 / 3} \Rightarrow 2^3=\frac{A}{9}\)

∴ A = 72 = number of nucleons.

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Question 36. The volume occupied by an atom is greater than the volume of the nucleus by a factor of about

1. 10¹⁰
2. 10¹⁵
3. 10²
4. 10⁵

Answer: 2. 10¹⁵

The average atomic size is of the order of 10^-10 m.

The nuclear radius is around1 fm = 10^-15 m

Hence, \(\frac{\text { volume of the atom }}{\text { volume of the nucleus }}=\frac{\frac{4}{3} \pi\left(10^{-10} \mathrm{~m}\right)^3}{\frac{4}{3} \pi\left(10^{-15} \mathrm{~m}\right)^3}=\left(10^5\right)^3=10^{15}\).

Question 37. When a deuterium is bombarded on an \({ }_8^{16} \mathrm{O}\) nucleus, an α-particle is emitted. The product nucleus is.

1. \({ }_7^{13} \mathrm{~N}\)
2. \({ }_5^{10} \mathrm{~B}\)
3. \({ }_4^9 \mathrm{Be}\)
4. \({ }_7^{14} \mathrm{~N}\)

Answer: 4. \({ }_7^{14} \mathrm{~N}\)

The given reaction is \({ }_1^2 \mathrm{H}+{ }_8^{16} \mathrm{O} \rightarrow{ }_2^4 \alpha+{ }_Z^A \mathrm{X}\)

where the product nucleus has the mass number A and the atomic number Z. Since the mass number is conserved,

2 +16 = 4 + A ⇒ A = 14.

The atomic number is conserved, so

l + 8 = 2 + Z

Z = 9 – 2 = 7.

Hence, the product nucleus is \({ }_7^{14} \mathrm{~N}\)

Question 38. In the reaction \(\mathrm{X}(\mathrm{n}, \alpha) \rightarrow{ }_3^7 \mathrm{Li}\), the element X is

1. \({ }_2^4 \mathrm{He}\)
2. \({ }_5^{10} \mathrm{~B}\)
3. \({ }_5^9 \mathrm{~B}\)
4. \({ }_4^{11} \mathrm{Be}\)

Answer: 2. \({ }_5^{10} \mathrm{~B}\)

The given nuclear reaction can be expressed as

⇒ \({ }_{\mathrm{Z}}^A \mathrm{X}+{ }_0^1 \mathrm{n} \rightarrow{ }_3^7 \mathrm{Li}+{ }_2^4 \mathrm{He}\)

Conserving the mass number,

A + l = 7 + 4 ⇒ A = 10.

Conserving the atomic number,

Z + 0 = 3 + 2 ⇒ Z = 5.

Thus, \({ }_Z^A \mathrm{X}={ }_5^{10} \mathrm{X}\)corresponds to the element boron \(\left({ }_5^{10} B\right)\)

Question 39. If a nuclear decay is expressed as \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+\beta^{+}+\mathrm{X}\), the unknown particle is a/an

1. Neutron
2. Antineutrino
3. Proton
4. Neutrino

Answer: 4. Neutrino

The beta-positive decay reaction is given as

⇒ \({ }_6^{11} \mathrm{C} \rightarrow{ }_5^{11} \mathrm{~B}+{ }_1^0 \beta+{ }_{\mathrm{Z}}^A \mathrm{X}\)

Conserving the mass number,

11 = 11 + 0 + A ⇒ A = 0.

Conserving the atomic number,

6 = 5 +1 + Z ⇒ Z = 0.

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With β⁺ activity, an unknown particle is emitted. This particle has a zero charge (Z = 0) and a zero mass (A = 0). Hence, the unknown particle is a neutrino.

Question 40. Complete the equation for the following fission process: \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{38}^{90} \mathrm{~S}+\ldots\)

1. \({ }_{54}^{143} \mathrm{Xe}+3{ }_0^1 \mathrm{n}\)
2. \({ }_{54}^{145} \mathrm{Xe}\)
3. \({ }_{57}^{142} \mathrm{Xe}\)
4. \({ }_{54}^{142} \mathrm{Xe}+{ }_0^1 \mathrm{n}\)

Answer: 1. \({ }_{54}^{143} \mathrm{Xe}+3{ }_0^1 \mathrm{n}\)

Let the unknown product be \({ }_Z^A \mathrm{X}\). Hence,

⇒ \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{38}^{90} \mathrm{Sr}+{ }_{\mathrm{Z}}^A \mathrm{X}\).

Conserving the mass number,

235 +1 = 90 + A => A = 146.

Conserving the atomic number,

92 + 0 = 38 + Z ⇒ Z = 54

Science \({ }_{54}^{146} \mathrm{Xe}\) is not given as the product, the product must be \({ }_{54}^{143} \mathrm{Xe}\) and \(3{ }_0^1 \mathrm{n}\), which gives A = 143 + 3 = 146 and Z = 54.

Question 41. A nucleus \({ }_n^m X\) emits one a- and two p-particles. The resulting nucleus is

1. \(\underset{n}{m-4} X\)
2. \({ }_{n-2}^{m-4} \mathrm{Y}\)
3. \({ }_{n-4}^{m-4} \mathrm{Z}\)
4. None of these

Answer: 1. \(\underset{n}{m-4} X\)

The given decay process is expressed as

⇒ \({ }_n^m \mathrm{X} \rightarrow{ }_2^4 \mathrm{He}+2{ }_{-1}^0 \beta\)

We can express the decay in the two steps shown below:

⇒ \({ }_n^m \mathrm{X} \rightarrow{ }_2^4 \mathrm{He}+{ }_{n-2}^{m-4} \mathrm{Y}\) (α-decay)

and \({ }_{n-2}^{m-4} \mathrm{Y} \rightarrow 2_{-1}^0 \beta+{ }_n^{m-4} \mathrm{Z}\) (β-decay)

Since \({ }_n^m \mathrm{X}\) and the final product \({ }_n^{m-4} \mathrm{Z}\) have the same atomic number (= n), X and Z may be the same element. Hence, the resulting nucleus is \({ }_n^{m-4} \mathrm{X}\).

Question 42. A nucleus at rest breaks into two fragments, which have their velocities in the ratio 2:1. What will be the ratio of their nuclear sizes (radii)?

1. \(2^{1 / 3}: 1\)
2. \(1: 2^{1 / 3}\)
3. \(3^{1 / 2}: 1\)
4. \(1: 3^{1 / 2}\)

Answer: 2. \(1: 2^{1 / 3}\)

Since the linear momentum is conserved,

⇒ \(\left|m_1 v_1\right|=\left|m_2 v_2\right|\)

Scince \(\frac{v_1}{v_2}=\frac{2}{1}\)

⇒ \(\frac{m_1}{m_2}=\frac{1}{2}=\frac{\frac{4}{3} \pi R_1^3 \rho}{\frac{4}{3} \pi R_2^3 \rho}\)

∴ \(\left(\frac{R_1}{R_2}\right)^3=\frac{1}{2} \Rightarrow R_1: R_2=1: 2^{1 / 3}\).

Question 43. An element A decays into another element C by the following two-step process: \(\mathrm{A} \rightarrow \mathrm{B}+{ }_2^4 \mathrm{He}; \quad \mathrm{B} \rightarrow \mathrm{C}+2 \mathrm{e}\) Then,

1. A and C are isotopes
2. A and C are isobars
3. A and B are isotopes
4. A and B are isobars

Answer: 1. A and C are isotopes

Let the element A have the mass number A and the atomic number Z.

So, in the first case,

⇒ \({ }_{\mathrm{Z}}^A \mathrm{~A} \rightarrow{ }_{\mathrm{Z}-2}^{A-4} \mathrm{~B}+{ }_2^4 \mathrm{He}\)

Similarly, in the second case,

⇒ \({ }_{Z-2}^{A-4} \mathrm{~B} \rightarrow{ }_Z^{A-4} \mathrm{C}+2{ }_{-1}^0 \mathrm{e}\)

Since the elements A and C have the same atomic number (= Z) but different mass numbers, they are isotopes of each other.

Question 44. What is the binding energy per nucleon in the \({ }_{26}^{56} \mathrm{Fe}\) nucleus? [Given that m\(\left({ }^{56} \mathrm{Fe}\right)\) = 55.936 u, mn = 1.00866 u and mp = 1.007274 u.]

1. 8.52 MeV
2. 10.56 MeV
3. 20.52 MeV
4. 50.5 MeV

Answer: 1. 8.52 MeV

The nucleus \({ }_{26}^{56} \mathrm{Fe}\) consists of 26 protons and 30 neutrons.

∴ 26 m_p = 26(1.007274 u) = 26.189124 u

and 30m_n = 30(1.00866 u) = 30.2598 u.

The total mass of the constituents = 56.448924 u.

Mass of the nucleus = m(\({ }_{26}^{56} \mathrm{Fe}\)) = 55.936 u.

∴ mass loss = ΔM = 0.513 u.

∴ the binding energy per nucleon is

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∴ \(\frac{\Delta M(931 \mathrm{MeV})}{56}=8.52 \mathrm{MeV}\)

Question 45. Consider the nuclear fission \({ }^{20} \mathrm{Ne} \rightarrow 2^4 \mathrm{He}+{ }^{12} \mathrm{C}\). Given that the binding energy per nucleon of \({ }^{20} \mathrm{Ne},{ }^4 \mathrm{He} \text { and }{ }^{12} \mathrm{C}\) are respectively 8.03 MeV, 7.07 MeV and 7.86 MeV. Identify the correct statement.

1. An energy of 3.6 MeV will be released.
2. An energy of 9.72 MeV will be absorbed.
3. An energy of 8.3 MeV will be released.
4. An energy of 11.9 MeV has to be supplied

Answer: 2. An energy of 9.72 MeV will be absorbed.

The Q-value of a nuclear reaction is given by

ΔQ = (BE of the products)-(BE of the reactants)

Energy is absorbed if ΔQ is negative and released if ΔQ is positive.

Hence, BE of \({ }^{20} \mathrm{Ne}\) = 20(8.03 MeV) =160.6 MeV,

BE of two \({ }^4 \mathrm{He}\) = 2(7.07 MeV)4 = 56.56 MeV

and BE of \({ }^{12} \mathrm{C}\) = 12(7.86 MeV) = 94.32 MeV.

∴ BE of all the products = 150.88 MeV.

∴ ΔQ =.150.88 MeV – 160.6 MeV = -9.72 MeV.

Hence, 9.72 MeV is the amount of energy absorbed.

Question 46. The ratio of the mass densities of the nuclei of \({ }^{40} \mathrm{Ca}\) and \({ }^{16} \mathrm{O}\) is close to

1. 3
2. 5
3. 1
4. 0.1

Answer: 3. 1

The density of the nuclear matter is constant and is independent of the size of the nucleus. Thus,

∴ \(\frac{R_{\mathrm{Ca}}}{R_{\mathrm{O}}}=1\)

Question 47. In a radioactive-decay chain, the initial nucleus is \({ }_{90}^{232} \mathrm{Th}\). In the end, there are six a-particles, and four which have been emitted. If the resulting nucleus is \({ }_{\mathrm{Z}}^A \mathrm{X}\) the values of A and Z are respectively

1. 202 and 80
2. 208 and 82
3. 200 and 81
4. 208 and 80

Answer: 2. 208 and 82

The nuclear reaction can be expressed as

⇒ \({ }_{90}^{232} \mathrm{Th} \rightarrow 6\left({ }_2^4 \mathrm{He}\right)+4{ }_{-1}^0 \beta+{ }_Z^A \mathrm{X}\)

Now, A + 4(0) + 6(4) = 232 ⇒ A = 208.

Also, Z + 4(-1) + 6(2) = 90 ⇒ Z = 82.

Question 48. The masses of various atomic particles are given below:

mp =1.0072 u, mn =1.0087 u, me = 0.00548 u, mv = 0 and md = 2.0141 u,

where p is a proton, n is a neutron, e is an electron, \(\bar{v}\) is an antineutrino and d is a deuteron. Which of the following processes is allowed by the conservation of momentum and energy?

1. n + n → a deuterium atom (with an electron bound to the nucleus)
2. \(\mathrm{p} \rightarrow \mathrm{n}+\overline{\mathrm{e}}+\overline{\mathrm{v}}\)
3. \(\overline{\mathrm{e}}+\mathrm{e} \rightarrow \gamma\)
4. n + p → d + γ

Answer: 4.

A nuclear reaction occurs when the mass of the products is less than the mass of the reactants. Let us consider the given options one by one.

1. n + n → a deuterium atom is energetically not feasible.
2. p → n + \(\overline{\mathrm{e}}+\overline{\mathrm{v}}\) is energetically not feasible.
3. e + \(\overline{\mathrm{e}}\) → γ is incorrect as the momentum is not conserved.
4. n + p → d + γ is energetically possible with the momentum conserved

Question 49. Given that mass of \({ }_3^7 \mathrm{Li}=1.0160 \mathrm{u}\), mass of \({ }_2^4 \mathrm{He}=4.0026 \mathrm{u}\) and mass of \({ }_1^1 \mathrm{H}=1.0070\). When 20 g of \({ }_3^7 \mathrm{Li}\) is converted into \({ }_2^4 \mathrm{He}\) by proton capture, the energy released is

1. 6.82 x 10⁵ kW h
2. 8 x 10⁶ kW h
3. 1.35 x l0⁶ kW h
4. 4.5 x 10⁵ kW h

Answer: 3. 1.35 x l0⁶ kW h

For the reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow 2{ }_2^4 \mathrm{He}\), we obtain.

⇒ \(m\left({ }_3^7 \mathrm{Li}\right)+m\left({ }_1^1 \mathrm{H}\right) \rightarrow 2 m\left({ }_2^4 \mathrm{He}\right)\).

Mass loss = Δm = [7.0160 +1.0070- 2(4.0026)] u = 0.0178 u.

For 20 g of \({ }_3^7 \mathrm{Li}\) the energy released will be

⇒ \(E=\left(\frac{20}{7} N_{\mathrm{A}}\right)(0.0178 \mathrm{u})\), where NA is the Avogadro constant

⇒ \(\frac{20}{7} \times\left(6 \times 10^{23} \mathrm{~mol}^{-1}\right) \times 0.0178 \mathrm{u}\)

⇒ \(\frac{20 \times 6 \times 10^{23} \times 0.0178 \times 10^9 \times 1.6 \times 10^{-19}}{7 \times 1000 \times 3600} \mathrm{~kW} \mathrm{~h}\),

∴ 1.35 x l0⁶ kW h.

Question 50. The radius (R) of a nucleus of mass number A can be estimated by the formula R = (1.3 x l0^-15 m)A^1/3. It follows that the mass density of the nuclear matter is of the order of (given that m_p = m_n ≈ 1.67 x l0^-27 kg)

1. 10³ kg m^-3
2. 10¹⁷ kg m^-3
3. 10²⁴ kg m^-3
4. 10¹⁰ kg m^-3

Answer: 2. 10¹⁷ kg m^-3

Density = \(\frac{\text { mass }}{\text { volume }}=\frac{m_{\mathrm{p}} A}{\frac{4}{3} \pi R^3}=\frac{m_{\mathrm{p}} A}{\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3}\)

⇒ \(\frac{3}{4 \pi} \cdot \frac{m_{\mathrm{p}}}{R_0^3}=\frac{3}{4 \pi} \cdot \frac{1.67 \times 10^{-27} \mathrm{~kg}}{\left(1.3 \times 10^{-15} \mathrm{~m}\right)^3}\)

⇒ \(\frac{3 \times 1.67}{4 \pi(1.3)^3} \times 10^{18} \mathrm{~kg} \mathrm{~m}^{-3}\)

∴ 1.81 x l0¹⁷ kg m^-3

Question 51. When the uranium nuclide \({ }_{92}^{235} \mathrm{U}\) is bombarded with a neutron, it generates the nuclide \({ }_{36}^{89} \mathrm{Kr}\), three neutrons and the nuclide

1. \({ }_{40}^{91} \mathrm{Zr}\)
2. \({ }_{36}^{101} \mathrm{Kr}\)
3. \({ }_{36}^{103} \mathrm{Kr}\)
4. \({ }_{56}^{144} \mathrm{Ba}\)

Answer: 4. \({ }_{40}^{91} \mathrm{Zr}\)

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⇒ \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{92}^{236} \mathrm{U} \rightarrow{ }_{36}^{89} \mathrm{Kr}+{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+3{ }_0^1 \mathrm{n}\)

Conserving the mass number,

235 + 1 = 89 + 3 + A ⇒ A =144.

Conserving the atomic number,

92 = 36 + Z ⇒ Z = 56.

Hence, the given nucleus \({ }_Z^A X\) is actually \({ }_{56}^{144} \mathrm{Be}\).

Question 52. The energy equivalent of 0.5 g of matter is

1. 4.5 x l0¹⁶ J
2. 4.5 x 10¹³ J
3. 0.5 x l0¹³ J
4. 1.5 x l0¹³ J

Answer: 2. 4.5 x 10¹³ J

From Einstein’s mass-energy equivalence,

E = mc² = (0.5 x 10^-3 kg)(3 x 108 m s^-1)²

= 4.5 x l0¹³ J