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NEET • PHYSICS • NUCLEI

Nuclei practice · Mixed difficulty

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31 Easy
31 of 210
Analyzing +4 | -1
Assertion: 13H{ }_1^3 {H} isotope does not undergo fusion of the type 13H+12H→{ }_1^3 {H}+{ }_1^2 {H} \rightarrow as it is rarely found in nature.
Reason: 13H{ }_1^3 {H} has half life of >>12> >12 years.
View answer and explanation
Correct option: B
If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion
Option B (If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the half-life of Tritium doesn't directly explain why it's rarely involved in fusion reactions of the type mentioned.
32 Easy
32 of 210
Application +4 | -1
The atomic mass of 6C12{ }_6 \mathrm{C}^{12} is 12.000000 u12.000000 \mathrm{~u} and that of 6C13{ }_6 \mathrm{C}^{13} is 13.003354 u.13.003354 \mathrm{~u}. The required energy to remove a neutron from 6C13,{ }_6 \mathrm{C}^{13}, if mass of neutron is 1.008665 u,1.008665 \mathrm{~u}, will be:
View answer and explanation
Correct option: D
4.95 MeV4.95 \mathrm{~MeV}
Option D (4.95 MeV4.95 \mathrm{~MeV}) is correct because it represents the energy equivalent of the mass difference between the reactants and products in the neutron removal process.
33 Medium
33 of 210
Understanding +4 | -1
Fpp,Fnn  and  Fnp{F_{pp}},{F_{nn}}\,\,\mathrm{and}\,\,{F_{np}} are the nuclear forces between proton-proton, neutron-neutron and neutron-proton respectively. Then relation between them is
View answer and explanation
Correct option: C
Fpp=Fnn=Fnp{F_{pp}} = {F_{nn}} = {F_{np}}
Option C (Fpp=Fnn=Fnp{F_{pp}} = {F_{nn}} = {F_{np}}) is correct because nuclear forces are charge independent and act equally between proton-proton, neutron-neutron, and neutron-proton pairs.
34 Medium
34 of 210
Application +4 | -1
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1\mathrm{2:1}. The ratio of their nuclear sizes will be
View answer and explanation
Correct option: D
1:21/31:{2^{1/3}}
Option D (1:21/31:{2^{1/3}}) is correct because the ratio of nuclear sizes is inversely proportional to the ratio of their velocities, and the radius is proportional to the cube root of the mass number.
35 Medium
35 of 210
Remembering +4 | -1
Curie is a unit of
View answer and explanation
Correct option: C
Radioactivity
Option C is correct because the Curie (Ci) is a unit used to measure the radioactivity of a substance, specifically the number of nuclear decays per unit time.
36 Medium
36 of 210
Understanding +4 | -1
In gamma ray emission from a nucleus
View answer and explanation
Correct option: B
There is no change in the proton number and the neutron number
The correct answer is that there is no change in the proton number and the neutron number because gamma emission involves the release of energy from the nucleus without changing its composition.
37 Medium
37 of 210
Remembering +4 | -1
Atomic reactor is based on
View answer and explanation
Correct option: A
Controlled chain reaction
Option A (Controlled chain reaction) is correct because atomic reactors use controlled nuclear fission to generate energy.
38 Medium
38 of 210
Application +4 | -1
For thorium A=232A = 232,Z=90Z = 90 at the end of some radioactive disintegration we obtain an isotope of lead with A=208A = 208 and Z=82Z = 82, then the number of emitted α  and  β\mathrm{\alpha }\,\,\mathrm{and}\,\,\mathrm{\beta } particles are
View answer and explanation
Correct option: C
α=6,β=4\mathrm{\alpha } = 6,\mathrm{\beta } = 4
Option C is correct because alpha decay reduces the mass number by 4 and the atomic number by 2, while beta decay leaves the mass number unchanged and increases the atomic number by 1. By setting up equations based on the changes in mass and atomic numbers, we can solve for the number of alpha and beta particles emitted.
39 Medium
39 of 210
Understanding +4 | -1
In which of the following decay, the element does not change
View answer and explanation
Correct option: C
γ\gamma-decay
Option C is correct because gamma decay involves the emission of a high-energy photon, which does not change the number of protons or neutrons in the nucleus, thus the element remains the same.
40 Medium
40 of 210
Analyzing +4 | -1
Identify the pair of physical quantities that have same dimensions:
View answer and explanation
Correct option: D
wave number and Avogadro number
Option A (velocity gradient and decay constant) is correct because both have dimensions of inverse time (T−1T^{-1}).
41 Easy
41 of 210
Understanding +4 | -1
Statement-I: Positive values of packing fraction implies a large value of binding energy.

Statement-II : The difference between the mass of the nucleus and the mass number of the nucleus is called packing fraction.
View answer and explanation
Correct option: C
Statement-I is true and Statement-II is false
Option C (Statement-I is true and Statement-II is false) is correct because a positive packing fraction indicates lower binding energy, and the packing fraction is related to mass defect.
42 Medium
42 of 210
Remembering +4 | -1
An electron is
View answer and explanation
Correct option: D
A lepton
An electron is a fundamental particle and belongs to the lepton family.
43 Medium
43 of 210
Understanding +4 | -1
A radioactive nucleus can decay simultaneously by two different processes which have decay constant λ1  and  λ2{\lambda _1}\,\,\mathrm{and}\,\,{\lambda _2}.The effective decay constant of the nuclide is λ,\lambda , where
View answer and explanation
Correct option: A
λ=λ1+λ2\lambda = {\lambda _1} + {\lambda _2}
Option A is correct because when a radioactive nucleus decays by two independent processes, the effective decay constant is the sum of the individual decay constants.
44 Medium
44 of 210
Analyzing +4 | -1
U238{U^{238}} decays into Th234T{h^{234}} by the emission of an α−\alpha - particle. There follows a chain of further radioactive decays, either by α−\alpha - decay or by β−\beta - decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. Which of the following stable nuclides is the end product of the U238{U^{238}} radioactive decay chain
View answer and explanation
Correct option: A
Pb206P{b^{206}}
Option A (Pb206P{b^{206}}) is correct because after a series of alpha and beta decays, Uranium-238 eventually transforms into Lead-206.
45 Easy
45 of 210
Application +4 | -1
From a newly formed radioactive substance (Half life 2  hours2\,\,hours), the intensity of radiation is 64\mathrm{64} times the permissible safe level. The minimum time after which work can be done safely from this source is
View answer and explanation
Correct option: B
12  hours12\,\,hours
Option B (12 hours12\,hours) is correct because it takes 6 half-lives for the radiation intensity to reduce to a safe level (64 times to 1 time). Since the half-life is 2 hours, the total time is 6 * 2 = 12 hours.
46 Medium
46 of 210
Analyzing +4 | -1
A radioactive element forms its own isotope after 3\mathrm{3} consecutive disintegrations. The particles emitted are
View answer and explanation
Correct option: B
2  β\mathrm{2}\,\,\mathrm{\beta }-particles - 1  α\mathrm{1}\,\,\mathrm{\alpha }-particle
Option B (2 β-particles - 1 α-particle) is correct because emission of an alpha particle reduces mass number by 4 and atomic number by 2, while emission of two beta particles increases atomic number by 2, resulting in an isotope of the original element.
47 Medium
47 of 210
Understanding +4 | -1
The radioactive nucleus 7N13_7{N^{13}} decays to 6C13_6{C^{13}} through the emission of
View answer and explanation
Correct option: D
Positron
The decay of 7N13_7{N^{13}} to 6C13_6{C^{13}} involves a decrease in atomic number by 1 and no change in mass number, indicating positron emission.
48 Medium
48 of 210
Understanding +4 | -1
An observer AA sees an asteroid with a radioactive element moving by at a speed =0.3  c= 0.3\,\,c and measures the radioactivity decay time to be TA{T_A}. Another observer BB is moving with the asteroid and measures its decay time as TB{T_B}. Then TA{T_A} and TB{T_B} are related as below
View answer and explanation
Correct option: A
TB<TA{T_B} < {T_A}
Time dilation occurs in special relativity, so the observer A will measure a longer decay time than observer B, hence TB<TA{T_B} < {T_A}.
49 Easy
49 of 210
Understanding +4 | -1
An element of mass M\mathrm{M} has Z\mathrm{Z} protons and N\mathrm{N} neutrons. Masses of proton and neutron are mpm_p and mnm_n respectively. Choose the correct relation among following options
View answer and explanation
Correct option: C
M<Zmp+NmnM < Z m_p+N m_n
Option C (M<Zmp+NmnM < Z m_p+N m_n) is correct because the mass of the nucleus is always less than the sum of the masses of its constituent protons and neutrons due to the binding energy.
50 Easy
50 of 210
Understanding +4 | -1
The momentum of an electron revolving in nth n^{\text {th }} orbit is given by: (Symbols have their usual meanings)
View answer and explanation
Correct option: A
nh2πr\frac{n h}{2 \pi r}
Option A is correct because the angular momentum of an electron in the nth orbit is quantized and given by L=nh2π=mvrL = n\frac{h}{2\pi} = mvr, so p=mv=nh2πrp = mv = \frac{nh}{2\pi r}.
51 Medium
51 of 210
Analyzing +4 | -1
Given the masses of various atomic particles mp=1.0072 u,    mn=1.0087  u,    me=0.000548  u,    mvˉ=0,    md=2.0141  u,{m_p} = 1.0072\,u,\,\,\,\,{m_n} = 1.0087\,\,u,\,\,\,\,{m_e} = 0.000548\,\,u,\,\,\,\,{m_{\bar v}} = 0,\,\,\,\,{m_d} = 2.0141\,\,u, Where pp = proton, nn = neutron, ee = electron vˉ=\bar v = antineutrino and dd = deuteron. Which of the following process is allowed by momentum and energy conservation?
View answer and explanation
Correct option: C
n+p→d+vˉn + p \to d + \bar v
Option B (p→n+e++vˉp \to n + {e^ + } + \bar v) is correct because it is the only option that violates energy conservation, making it not allowed.
52 Medium
52 of 210
Understanding +4 | -1
Which can pass through 20  cm20\,\,cm thickness of the steel
View answer and explanation
Correct option: C
γ−\gamma - rays
Option C (γ−\gamma - rays) is correct because gamma rays are high-energy photons with no mass or charge, allowing them to penetrate thick materials like steel.
53 Easy
53 of 210
Application +4 | -1
A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2:12: 1. After disintegration they will move :-
View answer and explanation
Correct option: A
In opposite directions with speed in the ratio of 1:21: 2 respectively
Option A (In opposite directions with speed in the ratio of 1:21: 2 respectively) is correct because of conservation of momentum. The lighter fragment will have twice the speed of the heavier fragment.
54 Medium
54 of 210
Understanding +4 | -1
Radio carbon dating is done by estimating in specimen the
View answer and explanation
Correct option: C
Ratio of amount of 14C6  to  12C6^{14}{\mathrm{C}_6}\,\,\mathrm{to}\,{\,^{12}}{\mathrm{C}_6} still present
Option C is correct because radiocarbon dating relies on measuring the ratio of 14C6^{14}{\mathrm{C}_6} to 12C6^{12}{\mathrm{C}_6} to determine the age of a specimen.
55 Medium
55 of 210
Application +4 | -1
After one alpha particle emission and one β\beta-particle emission from a nucleus
View answer and explanation
Correct option: D
Atomic number reduces by 1
Option D is correct because after alpha emission, the mass number decreases by 4 and the atomic number decreases by 2. After beta-minus emission, the mass number remains the same, and the atomic number increases by 1. Therefore, the net change is a decrease of 4 in mass number and a decrease of 1 in atomic number.
56 Medium
56 of 210
Remembering +4 | -1
In a material medium, when a positron meets an electron both the particles annihilate leading to the emission of two gamma ray photons. This process forms the basis of an important diagnostic procedure called
View answer and explanation
Correct option: B
PET
PET (Positron Emission Tomography) is the correct answer because it uses the annihilation of positrons and electrons to produce gamma rays for imaging.
57 Medium
57 of 210
Application +4 | -1
Minimum energy required to take out the only one electron from ground state of He+H{e^ + } is
View answer and explanation
Correct option: B
54.4  eV54.4\,\,eV
Option B (54.4 eV54.4 \,eV) is correct because the ionization energy of a hydrogen-like ion is proportional to the square of its atomic number.
58 Medium
58 of 210
Analyzing +4 | -1
The particle emitted in the decay of 92238U{ }_{92}^{238} \mathrm{U} to 92234U{ }_{92}^{234} \mathrm{U}
View answer and explanation
Correct option: B
1α1 \alpha and 2β2 \beta
Option B is correct because the mass number decreases by 4 (238 to 234) and the atomic number decreases by 2 (92 to 90) indicating the emission of an alpha particle, followed by two beta decays to restore the atomic number to 92.
59 Medium
59 of 210
Analyzing +4 | -1
If the binding energies of a deuteron and an alpha particle are 1.125  MeV\mathrm{1}\mathrm{.125}\,\,\mathrm{MeV} and 7.2  MeV\mathrm{7}\mathrm{.2}\,\,\mathrm{MeV}, respectively , then the more stable of the two is
View answer and explanation
Correct option: B
Alpha-particle
Option B (Alpha-particle) is correct because the binding energy per nucleon is higher for the alpha particle (7.2 MeV / 4 nucleons = 1.8 MeV/nucleon) than for the deuteron (1.125 MeV / 2 nucleons = 0.5625 MeV/nucleon).
60 Medium
60 of 210
Remembering +4 | -1
1\mathrm{1} curie represents
View answer and explanation
Correct option: C
3.7  ×  10103.7\,\, \times \,\,{10^{10}} disintegration per second
Option C (3.7 × 10103.7\, \times \,{10^{10}} disintegration per second) is correct because 1 curie is defined as the activity of a radioactive substance that undergoes 3.7 × 1010 disintegrations per second.
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