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NEET • PHYSICS • NUCLEI

Nuclei practice · Mixed difficulty

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181 Medium
181 of 210
Analyzing +4 | -1
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then:
View answer and explanation
Correct option: B
The helium nucleus has more kinetic energy than the thorium nucleus
The helium nucleus has more kinetic energy than the thorium nucleus. This is due to conservation of momentum and the lighter mass of the helium nucleus.
182 Medium
182 of 210
Remembering +4 | -1
The energy equivalent to a kilogram of matter is about
View answer and explanation
Correct option: B
1017  J{10^{17}}\,\,\mathrm{J}
Option B (1017 J{10^{17}}\,\mathrm{J}) is correct because it represents the energy equivalent calculated using Einstein's mass-energy equivalence formula, E=mc2.
183 Medium
183 of 210
Remembering +4 | -1
In helium nucleus, there are
View answer and explanation
Correct option: C
2  protons\mathrm{2}\,\,\mathrm{protons} and 2  neutrons\mathrm{2}\,\,\mathrm{neutrons}
Option C (2 protons\mathrm{2}\,\mathrm{protons} and 2 neutrons\mathrm{2}\,\mathrm{neutrons}) is correct because a helium nucleus consists of 2 protons and 2 neutrons.
184 Medium
184 of 210
Application +4 | -1
A radioactive nucleus of mass MM emits a photon of frequency vv and the nucleus recoils. The recoil energy will be
View answer and explanation
Correct option: C
h2v22Mc2\frac{{{h^2}{v^2}}}{{2M{c^2}}}
Option C (h2v22Mc2\frac{{{h^2}{v^2}}}{{2M{c^2}}}) is correct because the recoil energy can be derived from the conservation of momentum and energy, considering the emitted photon and the recoiling nucleus.
185 Easy
185 of 210
Understanding +4 | -1
In the following nuclear reaction x{x} stands for n→p+e−+xn \rightarrow p+e^{-}+x
View answer and explanation
Correct option: D
Antineutron
Option D is correct because in beta decay, a neutron decays into a proton, an electron, and an antineutrino to conserve lepton number.
186 Hard
186 of 210
Application +4 | -1
A radioactive substance with decay constant of 0.5/s0.5 / {s} is being produced at a constant rate of 50 nuclei per second. If there are no nuclei present initially, the time (in second) after which 25 nuclei will be present is:
View answer and explanation
Correct option: B
2 ln⁡ (4/3)2 \,\ln\, (4 / 3)
Option B (2 ln⁡ (4/3)2 \,\ln\, (4 / 3)) is correct because it solves the differential equation for the number of nuclei present, considering both production and decay.
187 Medium
187 of 210
Remembering +4 | -1
If λ\lambda is decay constant and NN the number of radioactive nuclei of an element , then the decay rate (RR) of that element is
View answer and explanation
Correct option: B
λN\lambda N
The decay rate is directly proportional to both the decay constant and the number of radioactive nuclei, hence the correct answer is λN\lambda N.
188 Medium
188 of 210
Understanding +4 | -1
Neutron decay in free space is given as follows
0n1→1H1+−1e0+  [ ]_0{n^1}{ \to _1}{\mathrm{H}^1} + { - _1}{e^0} + \,\,[\,] Then the parenthesis represents a
View answer and explanation
Correct option: C
Antineutrino
Neutron decay involves the emission of an antineutrino to conserve lepton number.
189 Medium
189 of 210
Application +4 | -1
In the reaction 12H+ 13  H→24He+ 01  n_1^2H + \,_1^3\,\,H \to _2^4He + \,_0^1\,\,n if the binding energies of 12H,13 H_1^2H,_1^3\,H and 24He_2^4He are respectively a,ba,b and cc (in MeVMeV), then the energy (in MeVMeV) released in this reaction is
View answer and explanation
Correct option: B
c−a−bc - a - b
Option B (c−a−bc - a - b) is correct because the energy released is the difference between the binding energy of the products and the reactants.
190 Medium
190 of 210
Application +4 | -1
The nucleus 48Cd115{}_{48}C{d^{115}} after two successive β−{\beta ^ - } decays will give
View answer and explanation
Correct option: D
50Cd115{}_{50}C{d^{115}}
Option D (50Cd115{}_{50}C{d^{115}}) is correct because beta-minus decay increases the atomic number by 1 while the mass number remains the same.
191 Easy
191 of 210
Remembering +4 | -1
Nueclius density:
View answer and explanation
Correct option: D
almost remain same
Option D ( almost remain same) is correct because the nucleus density is independent of the mass number.
192 Medium
192 of 210
Analyzing +4 | -1
Given below are two statements: one is labelled as Assertion A\mathbf{A} and the other is labelled as Reason R\mathbf{R}

Assertion A: The nuclear density of nuclides 510 B{ }_5^{10} \mathrm{~B}, 36Li,2656Fe,1020Ne{ }_3^6 \mathrm{Li},{ }_{26}^{56} \mathrm{Fe},{ }_{10}^{20} \mathrm{Ne} and 83209Bi{ }_{83}^{209} \mathrm{Bi} can be arranged as ρBiN>ρFeNN>ρNeNN>ρBN>ρLiN\rho_{\mathrm{Bi}}^{\mathrm{N}}>\rho_{\mathrm{Fe}_{\mathrm{N}}}^{\mathrm{N}}>\rho_{\mathrm{Ne}_{\mathrm{N}}}^{\mathrm{N}}>\rho_{\mathrm{B}}^{\mathrm{N}}>\rho_{\mathrm{Li}}^{\mathrm{N}}

Reason R\mathbf{R} : The radius RR of nucleus is related to its mass number AA as R=R0A1/3R=R_0 A^{1 / 3}, where R0R_0 is a constant.

In the light of the above statements, choose the correct answer from the options given below
View answer and explanation
Correct option: D
AA is false but RR is true
Option D (AA is false but RR is true) is correct because nuclear density is approximately constant for all nuclei, and the radius formula is correct.
193 Medium
193 of 210
Application +4 | -1
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)
View answer and explanation
Correct option: C
23.6 MeV
Option C (23.6 MeV) is correct because the energy released is the difference between the binding energy of the helium nucleus and the total binding energy of the two deuterons.
194 Medium
194 of 210
Analyzing +4 | -1
A radioactive nucleus ZAX_Z^AX undergoes spontaneous decay in the sequence
ZAX→Z−1B→Z−3C→Z−2D,_Z^AX{ \to _{Z - 1}}B{ \to _{Z - 3}}C{ \to _{Z - 2}}D, where ZZ is the atomic number of element XX. The possible decay particles in the sequence are
View answer and explanation
Correct option: A
β+,α,β−{\beta ^ + },\alpha ,{\beta ^ - }
Option A (β+,α,β−{\beta ^ + },\alpha ,{\beta ^ - }) is correct because it correctly identifies the decay particles emitted in the given sequence of transformations.
195 Medium
195 of 210
Analyzing +4 | -1
The half-life period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially both of them have the same number of atoms. Then
View answer and explanation
Correct option: C
Y will decay at a faster rate than X initially
Option A (5) is correct because setting the time constant of the RC circuit equal to the average life of the radioactive sample ensures the ratio of charge to activity remains constant.
196 Medium
196 of 210
Understanding +4 | -1
When the number of nucleons in nuclei increase, the binding energy per nucleon
View answer and explanation
Correct option: D
First increases and then decreases with increases of mass number
Option D is correct because the binding energy per nucleon initially increases as the mass number increases due to the strong nuclear force becoming more effective, but eventually decreases due to the increasing repulsive forces between protons.
197 Medium
197 of 210
Remembering +4 | -1
The example of nuclear fusion is
View answer and explanation
Correct option: B
Formation of helium from hydrogen
Option B is correct because nuclear fusion involves the combination of light nuclei, such as hydrogen isotopes, to form heavier nuclei like helium, releasing energy.
198 Hard
198 of 210
Application +4 | -1
Assuming that the mass of proton is nearly equal to mass of neutron the minimum kinetic energy in 101eV10^1 \mathrm{eV} of a neutron for inelastic head on collision with a ground state hydrogen atom at rest is -
View answer and explanation
Correct option: A
2
Option A (2) is correct because the minimum kinetic energy required for an inelastic collision is related to the energy difference between the ground state and the first excited state.
199 Medium
199 of 210
Analyzing +4 | -1
Assertion: Density of all the nuclei is same.

Reason: Radius of nucleus is directly proportional to the cube root ofmass number.
View answer and explanation
Correct option: B
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
Option B (Both Assertion and Reason are true and Reason is not the correct explanation of Assertion) is correct because both statements are factually correct, but the constant density doesn't directly arise from the radius formula.
200 Medium
200 of 210
Analyzing +4 | -1
A radioactive sample decays with an average-life of 20 ms20 \mathrm{~ms}. A capacitor of capacitance 100μF100 \mu \mathrm{F} is charged to some potential and then the plates are connected through a resistance RR. What should be the value of R/40R / 40 so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time?
View answer and explanation
Correct option: A
5
Option A (Moderator) is correct because heavy water is used to slow down neutrons in a nuclear reactor, acting as a moderator.
201 Medium
201 of 210
Application +4 | -1
An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8 : 27. The ratio of the radii of the nuclei (assumed to be spherical) is :
View answer and explanation
Correct option: C
3:23\mathrm{ }:\mathrm{ }2
Option C (3 : 2) is correct because the ratio of radii is inversely proportional to the cube root of the velocity ratio, derived from momentum conservation and the relationship between mass and radius.
202 Hard
202 of 210
Application +4 | -1
The energy released in the fusion of 2 kg2 \mathrm{~kg} of hydrogen deep in the sun is EH\mathrm{E}_{\mathrm{H}} and the energy released in the fission of 2 kg2 \mathrm{~kg} of 235U{ }^{235} \mathrm{U} is EU\mathrm{E}_{\mathrm{U}}. The ratio EHEU\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{U}}} is approximately :

(Consider the fusion reaction as 411H+2e−→24He+2v+6γ+26.7 M eV{4}_1^1 \mathrm{H}+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 \mathrm{v}+6 \gamma+26.7 \mathrm{~M~eV}, energy released in the fission reaction of 235U{ }^{235} \mathrm{U} is 200 M eV200 \mathrm{~M~eV} per fission nucleus and NA=6.023×1023\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23})
View answer and explanation
Correct option: C
7.62
Option C (7.62) is correct because it is the approximate ratio of energy released in hydrogen fusion to uranium fission, calculated using the given values.
203 Easy
203 of 210
Understanding +4 | -1
Assertion: In U235{U}_{\mathrm{235}} fission reaction neutrons are required to be slowed down.
Reason: The probability of capture of slow moving neutrons is high for U235{U}_{\mathrm{235}}.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion
Option A is correct because slow neutrons have a higher probability of being captured by U-235 nuclei, leading to nuclear fission, which is essential for sustaining a chain reaction in a nuclear reactor.
204 Medium
204 of 210
Understanding +4 | -1
Nuclear fission can be explained based on
View answer and explanation
Correct option: B
Liquid drop model
The liquid drop model provides a good analogy for understanding nuclear fission, where the nucleus is treated like a liquid drop that can deform and split.
205 Medium
205 of 210
Application +4 | -1
If a proton and anti-proton come close to each other and annihilate, how much energy will be released
View answer and explanation
Correct option: B
3  ×  10−10  J3\,\, \times \,\,{10^{ - 10}}\,\,J
Option B (3 × 10−10 J3\, \times \,{10^{ - 10}}\,J) is correct because the energy released is twice the rest mass energy of a proton converted to energy.
206 Easy
206 of 210
Application +4 | -1
If radius of the 1327Al{ }_{13}^{27} A l nucleus is taken to be RAl,R_{A l}, then the radius of 52125Te{ }_{52}^{125} {Te} nucleus is:
View answer and explanation
Correct option: C
53RAl\frac{5}{3} R_{A l}
Option C (53RAl\frac{5}{3} R_{Al}) is correct because the nuclear radius is proportional to the cube root of the mass number, so the ratio of radii is the cube root of the ratio of mass numbers.
207 Hard
207 of 210
Analyzing +4 | -1
The following deuterium reactions and corresponding reaction energies are found to occur
14N(d,p)15N,Q=8.53 MeV{ }^{14} {N}(d, p){ }^{15} {N}, Q=8.53 \,\mathrm{MeV}
15N(d,α)13C,Q=7.58 MeV{ }^{15} {N}(d, \alpha){ }^{13} {C}, Q=7.58 \,\mathrm{MeV}
13N(d,α)11B,Q=5.16 MeV{ }^{13} {N}(d, \alpha){ }^{11} {B}, Q=5.16\, \mathrm{MeV}
(The notation 14N(d,p)15N{ }^{14} {N}(d, p){ }^{15} {N} represent the reaction 14N+d→15N+p{ }^{14} {N}+d \rightarrow{ }^{15} {N}+p 24He=4.0026 amu;12H=2.014 amu;11H=1.0078 amu;n=1.0087 amu (1 amu=931 MeV).{ }_2^4 {He}=4.0026 \mathrm{~amu} ; { }_1^2 {H}=2.014 \,\mathrm{amu} ; { }_1^1 {H}=1.0078 \mathrm{~amu}; n=1.0087 \mathrm{~amu} \,(1 \mathrm{~a m u}=931 \mathrm{~MeV}). The QQ values of the reaction 11B(α,n)14N{ }^{11} {B}(\alpha, n){ }^{14} {N} is:
View answer and explanation
Correct option: C
0.05 MeV0.05 \mathrm{~MeV}
Option C is correct because by manipulating the given nuclear reactions and their Q-values, we can find the Q-value for the desired reaction, which is approximately 0.05 MeV.
208 Easy
208 of 210
Understanding +4 | -1
Choose the best option for binding energy/nucleon if nuclear energy is released in fission. Binding energy is
View answer and explanation
Correct option: C
larger for fission fragments than for parent nucleus
Option C (larger for fission fragments than for parent nucleus) is correct because nuclear fission releases energy when the binding energy per nucleon of the fission fragments is higher than that of the parent nucleus.
209 Medium
209 of 210
Analyzing +4 | -1
Following statements related to radioactivity are given below:
(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.
(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.
(C) Slope of the graph of loge (no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time (τ).\left( \tau \right).
(D) Product of decay constant (λ)\left( \lambda \right) and half-life time (T1/2)\left( {{T_{1/2}}} \right) is not constant.
Choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: C
(B) and (C) only
Option C (If assertion is true but reason is false) is correct because while binding energy generally increases with atomic mass number, the density of the nucleus is approximately constant.
210 Medium
210 of 210
Understanding +4 | -1
Decay constant of radium is λ\lambda. By a suitable process its compound radium bromide is obtained. The decay constant of radium bromide will be
View answer and explanation
Correct option: A
λ\lambda
Option A is correct because the decay constant is an intrinsic property of the radioactive nucleus and is not affected by chemical bonding or the formation of compounds.
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