JEE Physics Nuclei MCQs Set 18

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Chapter-wise Objective Questions: Nuclei

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Multiple Choice Questions

Question 1: The half-life of \( {}^{215}\text{At} \) is 100 microseconds. The time taken for its radioactivity to decay to 1/16 of its initial value is
(a) 400 microseconds
(b) 6.3 microseconds
(c) 40 microseconds
(d) 300 microseconds
Show Answer & Explanation

Answer: (a) 400 microseconds

Explanation:
1. 1/16 is 4 half-lives.
2. Time \( 4 imes100=400 \) microseconds.

Teacher's Note:
1. Count half-lives.

Question 2: When \( {}^{235}_{92}\text{U} \) undergoes fission, 0.1 percent of its original mass is changed into energy. The energy released if 1 kg of U-235 undergoes fission is
(a) \( 9\times10^{10} \) J
(b) \( 9\times10^{11} \) J
(c) \( 9\times10^{12} \) J
(d) \( 9\times10^{13} \) J
Show Answer & Explanation

Answer: (d) \( 9\times10^{13} \) J

Explanation:
1. \( \Delta m=0.001\times1=10^{-3} \) kg.
2. \( E=\Delta mc^2=10^{-3}\times9\times10^{16}=9\times10^{13} \) J.

Teacher's Note:
1. E = mc^2 with m in kg.

Question 3: A nucleus of binding energy B contains Z protons and N neutrons (\( M_p \), \( M_n \) are the masses of proton and neutron). The mass M(N, Z) of the nucleus is
(a) \( NM_n+ZM_p-Bc^2 \)
(b) \( NM_n+ZM_p+Bc^2 \)
(c) \( NM_n+ZM_p-B/c^2 \)
(d) \( NM_n+ZM_p+B/c^2 \)
Show Answer & Explanation

Answer: (c) \( NM_n+ZM_p-B/c^2 \)

Explanation:
1. Mass defect equals the binding energy divided by \( c^2 \): the nucleus is lighter than its parts.
2. So \( M=NM_n+ZM_p-B/c^2 \).

Teacher's Note:
1. Nuclear mass is less than the sum of its nucleon masses.

Question 4: N atoms of a radioactive element emit n alpha particles per second. The half-life of the element is
(a) \( \frac nN \) s
(b) \( \frac Nn \) s
(c) \( \frac{0.693N}{n} \) s
(d) \( \frac{0.693}{N} \) s
Show Answer & Explanation

Answer: (c) \( \frac{0.693N}{n} \) s

Explanation:
1. Activity \( n=\lambda N \), so \( \lambda=\frac nN \).
2. Half-life \( =\frac{0.693}{\lambda}=\frac{0.693N}{n} \) s.

Teacher's Note:
1. Activity = lambda x number of nuclei.

Question 5: A star initially has \( 10^{40} \) deuterons. It produces energy via \( {}^2_1\text{H}+{}^2_1\text{H}\to{}^3_1\text{H}+p \) and \( {}^2_1\text{H}+{}^3_1\text{H}\to{}^4_2\text{He}+n \). Masses: H-2 2.014 u, p 1.007 u, n 1.008 u, He-4 4.001 u. If the star radiates \( 10^{16} \) W, the deuteron supply is exhausted in a time of the order of
(a) \( 10^6 \) s
(b) \( 10^8 \) s
(c) \( 10^{12} \) s
(d) \( 10^{16} \) s
Show Answer & Explanation

Answer: (c) \( 10^{12} \) s

Explanation:
1. Net: 3 deuterons give He-4 + p + n. Mass defect \( 3 imes2.014-(4.001+1.007+1.008)=0.026 \) u = 24.2 MeV per 3 deuterons.
2. Total energy \( \frac{10^{40}}{3}\times24.2\ \text{MeV}=1.3\times10^{28} \) J; time \( =\frac{1.3\times10^{28}}{10^{16}}=1.3\times10^{12} \) s.

Teacher's Note:
1. Combine the two steps into one net reaction.

Question 6: Let M1 be the mass of a \( {}^{20}_{10}\text{Ne} \) nucleus and M2 the mass of a \( {}^{40}_{20}\text{Ca} \) nucleus. Which pair of statements is correct?
(a) \( M_2=2M_1 \) and \( M_1<10(m_n+m_p) \)
(b) \( M_2>2M_1 \) and \( M_1<10(m_n+m_p) \)
(c) \( M_2<2M_1 \) and \( M_1>10(m_n+m_p) \)
(d) \( M_2<2M_1 \) and \( M_1<10(m_n+m_p) \)
Show Answer & Explanation

Answer: (d) \( M_2<2M_1 \) and \( M_1<10(m_n+m_p) \)

Explanation:
1. Ca-40 has a larger binding energy per nucleon than Ne-20, so two Ne-20 nuclei would give more mass than one Ca-40: \( M_2<2M_1 \).
2. Every nucleus is lighter than its separate nucleons: \( M_1<10(m_n+m_p) \).

Teacher's Note:
1. Mass defect is greater for a more tightly bound system.

Question 7:
Assertion (A): The density of all nuclei is the same.
Reason (R): The radius of a nucleus is directly proportional to the cube root of its mass number.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Since \( R=R_0A^{1/3} \), volume is proportional to A and mass is proportional to A, so density is independent of A (R is true).
2. R is the reason for the constant density.

Teacher's Note:
1. Nuclear density is about 2.3 x 10^17 kg/m^3.

Question 8:
Assertion (A): Radioactive nuclei emit beta-minus particles.
Reason (R): Electrons exist inside the nucleus.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. A neutron changes into a proton, electron and antineutrino, so beta-minus emission occurs (A true).
2. Electrons do not exist inside the nucleus; they are created at emission (R false).

Teacher's Note:
1. Nucleus contains only protons and neutrons.

Question 9:
Assertion (A): A nucleus \( {}^A_Z\text{X} \) undergoes 2 alpha, 2 beta-minus and 2 gamma emissions; the daughter is \( {}^{A-8}_{Z-2}\text{Y} \).
Reason (R): In alpha decay the mass number decreases by 4 and the atomic number by 2. In beta decay the mass number is unchanged and the atomic number increases by 1.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Two alphas: A - 8, Z - 4. Two betas: Z + 2. Gamma: no change.
2. The daughter is \( {}^{A-8}_{Z-2}\text{Y} \), and R gives exactly the rules used.

Teacher's Note:
1. Gamma emission does not change A or Z.

Question 10:
Assertion (A): The mass number of a nucleus is always less than its atomic number.
Reason (R): The mass number of a nucleus may be equal to its atomic number.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Mass number is never smaller than the atomic number, so A is false.
2. It equals the atomic number for hydrogen-1, so R is true.

Teacher's Note:
1. A = Z + N.

Question 11:
Assertion (A): It is not possible to use chlorine-35 as the fuel for fusion energy.
Reason (R): The binding energy of chlorine-35 is too small.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Fusing chlorine-35 would give a nucleus beyond the iron peak, which releases no energy: A is true.
2. Chlorine-35 has a large binding energy per nucleon (about 8.5 MeV), so R is false.

Teacher's Note:
1. Fusion releases energy only for light nuclei.

Question 12:
Assertion (A): Cobalt-60 is useful in cancer therapy.
Reason (R): Cobalt-60 is a source of gamma radiation capable of killing cancerous cells.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Cobalt-60 emits energetic gamma rays that destroy cancer cells, which is exactly why it is useful.

Teacher's Note:
1. Gamma sources are used in radiotherapy.

Question 13: A rock contains 4.20 mg of U-238 and 2.135 mg of Pb-206. Assume all the lead came from uranium decay (half-life of U-238 is \( 4.47\times10^9 \) y). The number of U-238 atoms the rock now contains is
(a) \( 1.06\times10^{19} \)
(b) \( 2.2\times10^{18} \)
(c) \( 2.12\times10^{19} \)
(d) \( 5.3\times10^{18} \)
Show Answer & Explanation

Answer: (a) \( 1.06\times10^{19} \)

Explanation:
1. \( N=\frac{4.20\times10^{-3}}{238}\times6.022\times10^{23}=1.06\times10^{19} \).

Teacher's Note:
1. N = (mass / molar mass) x Avogadro number.

Question 14: For the same rock (original U-238 atoms \( 1.69\times10^{19} \), present \( 1.06\times10^{19} \), half-life \( 4.47\times10^9 \) y), the age of the rock is
(a) \( 2.50\times10^9 \) y
(b) \( 2.80\times10^9 \) y
(c) \( 2.98\times10^9 \) y
(d) \( 4.80\times10^9 \) y
Show Answer & Explanation

Answer: (c) \( 2.98\times10^9 \) y

Explanation:
1. \( t=\frac{T}{\ln2}\ln\frac{N_0}{N}=\frac{4.47\times10^9}{0.693}\ln\frac{1.69}{1.06} \).
2. This gives \( 6.45\times10^9\times0.4624=2.98\times10^9 \) y.

Teacher's Note:
1. Use N = N0 exp(-lambda t).

Question 15: Two deuteron nuclei in a fusion reactor core each have kinetic energy 1.5kT and approach head-on to a separation of \( 4\times10^{-15} \) m, neglecting other interactions. Taking \( \frac{e^2}{4\pi\epsilon_0}=1.44\times10^{-9} \) eV m and \( k=8.6 imes10^{-5} \) eV/K, the required temperature T lies in the range
(a) \( 1.0\times10^9 \) K to \( 2.0\times10^9 \) K
(b) \( 2.0\times10^9 \) K to \( 3.0\times10^9 \) K
(c) \( 3.0\times10^9 \) K to \( 4.0\times10^9 \) K
(d) \( 4.0\times10^9 \) K to \( 5.0\times10^9 \) K
Show Answer & Explanation

Answer: (a) \( 1.0\times10^9 \) K to \( 2.0\times10^9 \) K

Explanation:
1. Coulomb energy at 4 fm: \( \frac{1.44\times10^{-9}}{4\times10^{-15}}=3.6\times10^5 \) eV.
2. Total kinetic energy of both is \( 3kT \), so \( T=\frac{3.6\times10^5}{3\times8.6\times10^{-5}}=1.4\times10^9 \) K.

Teacher's Note:
1. Both nuclei carry kinetic energy 1.5kT.

Question 16: Which of these designs of a D-D fusion reactor is most promising by the Lawson criterion (the product of density n and confinement time must exceed \( 5\times10^{14} \) s/cm\( ^3 \))?
(a) deuteron density \( 2.0\times10^{12} \) cm\( ^{-3} \), confinement time \( 5.0\times10^{-3} \) s
(b) deuteron density \( 8.0\times10^{14} \) cm\( ^{-3} \), confinement time \( 9.0\times10^{-1} \) s
(c) deuteron density \( 4.0\times10^{23} \) cm\( ^{-3} \), confinement time \( 1.0\times10^{-11} \) s
(d) deuteron density \( 1.0\times10^{24} \) cm\( ^{-3} \), confinement time \( 4.0\times10^{-12} \) s
Show Answer & Explanation

Answer: (b) deuteron density \( 8.0\times10^{14} \) cm\( ^{-3} \), confinement time \( 9.0\times10^{-1} \) s

Explanation:
1. Compute n times t: (a) \( 10^{10} \); (b) \( 7.2\times10^{14} \); (c) \( 4\times10^{12} \); (d) \( 4\times10^{12} \).
2. Only (b) exceeds \( 5\times10^{14} \).

Teacher's Note:
1. Lawson number = n x t.

Question 17: In the same He+ ions populated in n = 4, the wavelength of light emitted in the visible region (by the 4 to 3 transition) is nearly
(a) \( 6.5\times10^{-7} \) m
(b) \( 5.6\times10^{-7} \) m
(c) \( 4.8\times10^{-7} \) m
(d) \( 4.0\times10^{-7} \) m
Show Answer & Explanation

Answer: (c) \( 4.8\times10^{-7} \) m

Explanation:
1. \( \frac1\lambda=1.097\times10^7\times4\left(\frac19-\frac1{16}\right)=2.13\times10^6 \) per m.
2. So \( \lambda=4.7\times10^{-7} \) m, closest to \( 4.8\times10^{-7} \) m.

Teacher's Note:
1. The 4 to 2 and 4 to 1 lines are in the ultraviolet.

Question 18: In the mixture, the ratio of the kinetic energy of the n = 2 electron in the H atom to that in the He+ ion is
(a) \( \frac14 \)
(b) \( \frac12 \)
(c) 1
(d) 2
Show Answer & Explanation

Answer: (a) \( \frac14 \)

Explanation:
1. \( K=\frac{13.6Z^2}{n^2} \): H gives 3.4 eV and He\( ^+ \) gives 13.6 eV.
2. Ratio = 1/4.

Teacher's Note:
1. K is proportional to Z^2.

Question 19: In beta decay \( n\to p+e^-+\bar\nu_e \) of a neutron at rest, the maximum kinetic energy of the electron is \( 0.8\times10^6 \) eV (proton recoil energy negligible, antineutrino taken as massless). The maximum energy of the antineutrino is
(a) zero
(b) much less than \( 0.8\times10^6 \) eV
(c) nearly \( 0.8\times10^6 \) eV
(d) much larger than \( 0.8\times10^6 \) eV
Show Answer & Explanation

Answer: (c) nearly \( 0.8\times10^6 \) eV

Explanation:
1. The energy released is shared between the electron and antineutrino.
2. When the electron has almost zero kinetic energy, the antineutrino carries nearly all of it, about \( 0.8\times10^6 \) eV.

Teacher's Note:
1. Total decay energy is fixed.

Question 20: A sample of iodine-128 shows the decay rate falling from 392.2 counts/s at t = 4 min to 161.4 counts/s at t = 36 min (and to 1.00 counts/s at t = 218 min). The half-life is about
(a) 12.5 min
(b) 25 min
(c) 50 min
(d) 100 min
Show Answer & Explanation

Answer: (b) 25 min

Explanation:
1. \( \lambda=\frac1{32}\ln\frac{392.2}{161.4}=\frac{0.888}{32}=0.0277 \) per min.
2. Half-life \( =\frac{0.693}{0.0277}=25 \) min (the 4 to 218 min data give the same value).

Teacher's Note:
1. Use A = A0 exp(-lambda t) between two readings.

Question 21: It is proposed to use D-D fusion \( {}^2_1\text{H}+{}^2_1\text{H}\to{}^4_2\text{He} \) in a 200 MW (output) reactor using 25 percent of the fusion energy. Masses: H-2 2.0141 u, He-4 4.0026 u. The deuterium fuel needed per day is nearly
(a) 30 g
(b) 60 g
(c) 120 g
(d) 240 g
Show Answer & Explanation

Answer: (c) 120 g

Explanation:
1. Mass defect \( 2 imes2.0141-4.0026=0.0256 \) u = 23.8 MeV per reaction; fusion power needed \( 200/0.25=800 \) MW.
2. Energy per day \( 8\times10^8\times86400=6.9\times10^{13} \) J; reactions \( =\frac{6.9\times10^{13}}{3.8\times10^{-12}}=1.8\times10^{25} \), using \( 3.6\times10^{25} \) deuterium atoms, i.e. about 120 g.

Teacher's Note:
1. Each reaction consumes two deuterium nuclei.

Question 22: The disintegration rate of a sample is 4750 per minute at some instant and 2700 per minute five minutes later. The half-life is
(a) 3.1 min
(b) 6.1 min
(c) 12.2 min
(d) 9.2 min
Show Answer & Explanation

Answer: (b) 6.1 min

Explanation:
1. \( \lambda=\frac15\ln\frac{4750}{2700}=0.1129 \) per min.
2. Half-life \( =\frac{0.693}{0.1129}=6.1 \) min.

Teacher's Note:
1. Use the ratio of rates.

Question 23: A reactor using U-235 (200 MeV per fission) has 10 percent efficiency and produces 1000 MW. If it is to function for 10 years, the total mass of uranium required is about
(a) \( 3.8\times10^4 \) kg
(b) \( 3.8\times10^3 \) kg
(c) \( 3.8\times10^2 \) kg
(d) \( 3.8\times10^5 \) kg
Show Answer & Explanation

Answer: (a) \( 3.8\times10^4 \) kg

Explanation:
1. Thermal power \( =10^{10} \) W; energy in 10 years \( =10^{10}\times3.156\times10^8=3.16\times10^{18} \) J.
2. Fissions \( =\frac{3.16\times10^{18}}{3.2\times10^{-11}}=9.9\times10^{28} \) , mass \( =\frac{9.9\times10^{28}}{6.022\times10^{23}}\times235\ \text{g}=3.8\times10^4 \) kg.

Teacher's Note:
1. Efficiency 10 percent means thermal power is ten times the output.

Question 24: The ratio of U-235 to U-238 in natural uranium today is 0.0072. Their half-lives are \( 7.04\times10^8 \) y and \( 44.7\times10^8 \) y. The ratio \( 2\times10^9 \) years ago was about
(a) 3.8 percent
(b) 0.72 percent
(c) 7.2 percent
(d) 1.4 percent
Show Answer & Explanation

Answer: (a) 3.8 percent

Explanation:
1. Going back in time, each isotope was more abundant by \( e^{\lambda t} \).
2. \( \text{ratio}=0.0072\,e^{(\lambda_{235}-\lambda_{238})t} \) with \( \lambda_{235}-\lambda_{238}=(9.85-1.55)\times10^{-10}=8.3\times10^{-10} \) per y, giving \( e^{1.66}=5.25 \) and \( 0.0378 \), about 3.8 percent.

Teacher's Note:
1. The shorter-lived isotope was proportionally more abundant earlier.

Question 25: In the proton-proton mechanism that powers the Sun, 26.7 MeV is released for each event (four protons forming one He-4). The power of the Sun is \( 3.9\times10^{26} \) W. The rate at which hydrogen is consumed is about
(a) \( 6.1\times10^{11} \) kg/s
(b) \( 6.1\times10^{9} \) kg/s
(c) \( 1.5\times10^{11} \) kg/s
(d) \( 2.4\times10^{12} \) kg/s
Show Answer & Explanation

Answer: (a) \( 6.1\times10^{11} \) kg/s

Explanation:
1. Events per second \( =\frac{3.9\times10^{26}}{26.7\times1.6\times10^{-13}}=9.1\times10^{37} \).
2. Each uses 4 protons: \( 9.1\times10^{37}\times4\times1.67\times10^{-27}=6.1\times10^{11} \) kg/s.

Teacher's Note:
1. Four protons per event.

Question 26: Using atomic masses \( {}^{25}\text{Al}=24.990432 \) u and \( {}^{25}\text{Mg}=24.985839 \) u, the Q-value of the positron decay \( {}^{25}\text{Al}\to{}^{25}\text{Mg}+e^++\nu \) is (1 u = 931.5 MeV, electron rest energy 0.511 MeV)
(a) 3.25 MeV
(b) 4.28 MeV
(c) 1.02 MeV
(d) 2.14 MeV
Show Answer & Explanation

Answer: (a) 3.25 MeV

Explanation:
1. For positron decay with atomic masses, two electron masses must be subtracted: \( Q=(\Delta m)c^2-2m_ec^2 \).
2. \( \Delta m=0.004593 \) u = 4.278 MeV, so \( Q=4.278-1.022=3.26 \) MeV.

Teacher's Note:
1. The daughter atom has one more electron than the parent's nucleus supports.

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