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JEE • PHYSICS • NUCLEI

Nuclei practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 380 questions
31 Medium
31 of 380
Application +4 | -1
Half life of a radioactive element is 10\mathrm{10} days. The time during which quantity remains 1/10\mathrm{1/10} of initial mass will be
View answer and explanation
Correct option: C
33\mathrm{33} days
Option C (33 days) is the closest to the correct answer because after approximately 3.3 half-lives, the quantity of the radioactive element will be reduced to 1/10 of its initial mass.
32 Medium
32 of 380
Understanding +4 | -1
A nuclear decay is expressed as C611→B511+β++XC_6^{11} \to B_5^{11} + {\beta ^ + } + X. Then the unknown particle X is
View answer and explanation
Correct option: D
Neutrino
Option D (Neutrino) is correct because in positive beta decay, a neutrino is emitted to conserve lepton number.
33 Medium
33 of 380
Application +4 | -1
During mean life of a radioactive element, the fraction that disintegrates is
View answer and explanation
Correct option: C
e−1e\frac{{e - 1}}{e}
Option C (e−1e\frac{{e - 1}}{e}) is correct because it represents the fraction of radioactive material that has decayed after one mean lifetime.
34 Medium
34 of 380
Analyzing +4 | -1
The half-life period of radioactive element xx is same as the mean life time of another radioactive element yy Initially they have the same number of atoms. Then,
View answer and explanation
Correct option: B
y will decay faster than x
Option B is correct because the decay rate is proportional to the decay constant, and since the half-life of X is equal to the mean life of Y, Y has a larger decay constant and thus decays faster.
35 Medium
35 of 380
Application +4 | -1
In the decay 64Cu→64Ni+e++v{ }^{64} {Cu} \rightarrow{ }^{64} {Ni}+e^{+}+{v}, the maximum kinetic energy carried by the positron is found to be 0.650 MeV0.650 \mathrm{~MeV}.

(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?0.150 \mathrm{~MeV}?
(b) What is the momentum of this neutrino in kg m s−1?\mathrm{kg} \mathrm{~m} \mathrm{~s}^{-1}?

Use the formula applicable to a photon.
View answer and explanation
Correct option: A
(a) 500 keV;500 \mathrm{~keV}; (b) 2.67×10−22 kg m s−12.67 \times 10^{-22} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
Option A ((a) 500 keV;500 \mathrm{~keV}; (b) 2.67×10−22 kg m s−12.67 \times 10^{-22} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}) is correct because the neutrino energy is the difference between the maximum positron energy and the given positron kinetic energy, and the momentum is calculated using the photon-like relationship.
36 Medium
36 of 380
Application +4 | -1
If mass of U235=235.12142amuU^{235}=235.12142 \mathrm{amu}, mass of U236=1.008665amu\mathrm{U}^{236}=1.008665 \mathrm{amu}, then the energy required to remove one neutron from the nucleus U236\mathrm{U}^{236} is nearly about -
View answer and explanation
Correct option: B
6.5MeV6.5 \mathrm{MeV}
Option B (6.5MeV6.5 \mathrm{MeV}) is correct because the energy required to remove a neutron is equal to the mass defect multiplied by the energy equivalent of 1 amu (931 MeV).
37 Medium
37 of 380
Understanding +4 | -1
Nuclear fission experiments show that the neutrons split the uranium nuclei into two fragments of about same size. This process is accompanised by the emission of several
View answer and explanation
Correct option: C
Neutrons
Option C (Neutrons) is correct because nuclear fission is accompanied by the emission of several neutrons, which sustain the chain reaction.
38 Medium
38 of 380
Remembering +4 | -1
Atomic nuclei having same mass number but difference atomic numbers are called
View answer and explanation
Correct option: A
Isobars
Option A (Isobars) is correct because isobars are defined as atomic nuclei having the same mass number but different atomic numbers.
39 Easy
39 of 380
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 (nh2πr\frac{n h}{2 \pi r}) is correct because the angular momentum is quantized, and linear momentum is derived from it.
40 Medium
40 of 380
Remembering +4 | -1
Modern treatment method P.E.T. is besed on !
View answer and explanation
Correct option: B
Positron emission
Option B (Positron emission) is correct because P.E.T. stands for Positron Emission Tomography.
41 Medium
41 of 380
Remembering +4 | -1
In nuclear fusion process, heat energy released is
View answer and explanation
Correct option: A
26.7 MeV26.7 \, \mathrm{MeV}
Option A (26.7 MeV26.7 \, \mathrm{MeV}) is correct because it's the approximate energy released during the fusion of light nuclei into heavier ones, specifically in reactions like the proton-proton chain.
42 Medium
42 of 380
Application +4 | -1
An unstable element is produced in a nuclear reactor at a constant rate. If its half life is 100 years, how much time in years is required to produce 50%50 \% of the equilibrium quantity?
View answer and explanation
Correct option: A
100
Option A (100) is correct because the time required to produce 50% of the equilibrium quantity is equal to the half-life of the unstable element.
43 Medium
43 of 380
Application +4 | -1
Calculate the disintegration energy Q for the fission of  4298M{ ~}_{42}^{98} \mathrm{M} o into two equal fragments, 2149 Sc{ }_{21}^{49} \mathrm{~S} c. If QQ turns out to be positive, explain why this process does not occur spontaneously. Given that :

m(4298Mo)=97.90541 amum\left({ }_{42}^{98} \mathrm{Mo}\right) =97.90541 \mathrm{~amu}

m(2149Sc)=48.95002 amum\left({ }_{21}^{49} \mathrm{Sc}\right) =48.95002 \mathrm{~amu}

mn=1.00867 amum_n =1.00867 \mathrm{~amu}.
View answer and explanation
Correct option: B
944.6 MeV944.6 \mathrm{~MeV}
Option B (944.6 MeV944.6 \text{{ MeV}}) is correct because it is the disintegration energy Q calculated using the mass defect and the conversion factor from amu to MeV.
44 Medium
44 of 380
Understanding +4 | -1
For the radioactive nuclei that undergo either α\alpha or β\beta-decay, which one of the following cannot occur?
View answer and explanation
Correct option: B
Isotope of original nucleus
Option B is correct because alpha and beta decay change the atomic number, thus an isotope of the original nucleus cannot be formed.
45 Medium
45 of 380
Understanding +4 | -1
Two nucleons are at a separation of 1 fm\mathrm{1 }\,\mathrm{fm}. The net force between them is F1{\mathrm{F}_\mathrm{1}} if both neutrons, F2{F_2} if both are protons, and F3{F_3} if one is a proton and the other is a neutron.
View answer and explanation
Correct option: C
F1=F3>F2{F_1} = {F_3} > {F_2}
Option C is correct because the nuclear force is approximately the same between any two nucleons (neutron-neutron, proton-proton, or proton-neutron), but the electrostatic repulsion between protons reduces the net force.
46 Medium
46 of 380
Understanding +4 | -1
Read the following statements:

(A) Volume of the nucleus is directly proportional to the mass number.

(B) Volume of the nucleus is independent of mass number.

(C) Density of the nucleus is directly proportional to the mass number.

(D) Density of the nucleus is directly proportional to the cube root of the mass number.

(E) Density of the nucleus is independent of the mass number.
Choose the correct option from the following options
View answer and explanation
Correct option: B
(A)(\mathrm{A}) and (E)(\mathrm{E}) only
The correct answer is B ((A) and (E) only) because the volume of the nucleus is proportional to the mass number, and the density is independent of it.
47 Easy
47 of 380
Application +4 | -1
r1   and   r2{r_1}\,\,\,{\text{and}}\,\,\,{r_2} are the radii of atomic nuclei of mass number 125 and 216 respectively. The ratio of r1r2\frac{{{r_1}}}{{{r_2}}} is
View answer and explanation
Correct option: B
56\frac{5}{6}
The correct answer is B (56\frac{5}{6}) because the radius of a nucleus is proportional to the cube root of its mass number, and the ratio of the radii is the cube root of the ratio of the mass numbers.
48 Medium
48 of 380
Analyzing +4 | -1
Fpe{\mathbf{F}_{pe}} represents electrical force on proton due to electron and Fep{\mathbf{F}_{ep}} on electron due to proton in a hydrogen atom.Similarly Fpe{\mathbf{F}_{pe}} represents the gravitational force on proton due to electron and Fep{\mathbf{F}_{ep}} the corresponding force on electron due to proton. Which of the following is not true?
View answer and explanation
Correct option: D
FPe+F′ Pe=0{\mathbf{F}_{Pe}} + \mathbf{F'}{\,_{Pe}} = 0
Option D (FPe+F′ Pe=0{\mathbf{F}_{Pe}} + \mathbf{F'}{\,_{Pe}} = 0) is not true because the electrical and gravitational forces between the proton and electron are not necessarily equal and opposite.
49 Medium
49 of 380
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 C (Y will decay at a faster rate than X initially) is correct because the decay constant of Y is greater than that of X, leading to a faster initial decay rate.
50 Medium
50 of 380
Remembering +4 | -1
Which of the following has the mass closest in value to that of the positron
1  a.m.u=931  MeV1\,\,a.m.u = 931\,\,MeV
View answer and explanation
Correct option: B
Electron
Option B (Electron) is correct because a positron is the antiparticle of an electron, and thus has the same mass as an electron.
51 Medium
51 of 380
Remembering +4 | -1
The density of uranium is of the order of
View answer and explanation
Correct option: A
1020  kgm−3{10^{20}}\,\,\mathrm{kg}{\mathrm{m}^{ - 3}}
Option D (1017 kg/m−3{10^{17}} \,\mathrm{kg}/\mathrm{m}^{ - 3}) is the closest order of magnitude to the actual density of the nucleus.
52 Medium
52 of 380
Understanding +4 | -1
Nuclear binding energy is equivalent to
View answer and explanation
Correct option: D
Mass defect of nucleus
Option D (Mass defect of nucleus) is correct because the nuclear binding energy is equivalent to the mass defect of the nucleus multiplied by c² (from E=mc²).
53 Medium
53 of 380
Understanding +4 | -1
Ionization power and penetration range of radioactive radiation increases in the order
View answer and explanation
Correct option: B
γ,β,α\mathrm{\gamma },\mathrm{\beta },\mathrm{\alpha } and α,β,γ\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma } respectively
Option D is correct because alpha particles have the highest ionization power and the lowest penetration range, while gamma rays have the lowest ionization power and the highest penetration range.
54 Medium
54 of 380
Application +4 | -1
The half-life of radium is 1600 years. After how many years 25%25 \% of a radium block remains undecayed?
View answer and explanation
Correct option: C
32003200 years
Option C (32003200 years) is correct because it accurately calculates the time required for the radium block to decay to 25% of its original amount, using the concept of half-life.
55 Medium
55 of 380
Application +4 | -1
A 280 days old radioactive substance shows an activity of 6000 dps, 140 days later it’s activity becomes 3000 dps. What was its initial activity.
View answer and explanation
Correct option: B
24000 dps
Option B (24000 dps) is correct because the initial activity is calculated by using the half-life formula and the given activities at different times.
56 Medium
56 of 380
Application +4 | -1
The radius of germanium (Ge\mathrm{Ge}) nuclide is measured to be twice the radius of 49Be{}_4^9\mathrm{Be}. The number of nucleons in Ge\mathrm{Ge} are
View answer and explanation
Correct option: D
72
Option D (72) is correct because the number of nucleons is calculated using the relationship between nuclear radius and mass number.
57 Medium
57 of 380
Remembering +4 | -1
Average binding energy of a nucleon inside an atomic nucleus is about
View answer and explanation
Correct option: A
8 MeV8 \, {MeV}
Option A (8 MeV8 \, {MeV}) is correct because the average binding energy per nucleon in an atomic nucleus is approximately 8 MeV.
58 Medium
58 of 380
Analyzing +4 | -1
Assertion: All natural radioactive elements are ultimately converted to lead.

Reason: All the elements above lead are unstable.
View answer and explanation
Correct option: B
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion.
Option B is correct because all natural radioactive elements decay until they reach a stable element, and lead is a stable element, but not all elements above lead are unstable.
59 Medium
59 of 380
Remembering +4 | -1
One Becquerel is defined as
View answer and explanation
Correct option: A
1\mathrm{1} disintegration per sec
Option A (1 disintegration per sec) is correct because the Becquerel (Bq) is the SI unit of radioactivity, defined as one disintegration per second.
60 Medium
60 of 380
Understanding +4 | -1
In β+\beta^{+}emission what happens to upquarks
View answer and explanation
Correct option: B
Is converted into down-quarks
Option B (Is converted into down-quarks) is correct because in β+\beta^+ decay, an up quark transforms into a down quark.
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