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JEE • PHYSICS • NUCLEI

Nuclei practice · Mixed difficulty

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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 271-300 of 380 questions

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271 Medium
271 of 380
Application +4 | -1
The binding energy per nucleon of O16{O^{16}} is 7.97  MeV7.97\,\,MeV and that of O17{O^{17}} is 7.75  MeV7.75\,\,MeV. The energy (in MeVMeV) required to remove a neutron from O17{O^{17}} is
View answer and explanation
Correct option: C
4.23
Option C (4.23 MeV) is correct because the energy required to remove a neutron is the difference in binding energies of O-17 and O-16.
272 Medium
272 of 380
Remembering +4 | -1
The example of nuclear fusion is
View answer and explanation
Correct option: B
Formation of helium from hydrogen
Option B (Formation of helium from hydrogen) is correct because nuclear fusion involves combining light nuclei to form a heavier nucleus, such as the fusion of hydrogen isotopes to form helium.
273 Medium
273 of 380
Application +4 | -1
In an α−\alpha - decay, the kinetic energy of α−\alpha - particle is 48 MeV and Q value of the reaction is 50 MeV. The mass number of the mother nucleus is [assume that daughter nucleus is in ground state]
View answer and explanation
Correct option: B
100
Option B (100) is correct because the mass number of the mother nucleus can be calculated using the kinetic energy of the alpha particle and the Q value of the reaction.
274 Easy
274 of 380
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 it correctly applies the formula for nuclear radius in terms of mass number.
275 Medium
275 of 380
Application +4 | -1
If 10%10 \% of a material decays in 5 days, then the amount of original material left after 20 days is approximately
View answer and explanation
Correct option: B
65%65 \%
Option B (65%) is correct because the remaining material is calculated using the exponential decay formula after determining the decay constant.
276 Medium
276 of 380
Application +4 | -1
How much mass of U235U^{235} is required per sec to operate the nuclear reactor generating power 100 kW100 \mathrm{~kW}.
View answer and explanation
Correct option: A
1.2×10−61.2 \times 10^{-6}
Option A (1.2×10−61.2 \times 10^{-6}) is correct because it is the mass of U-235 required per second to generate 100 kW of power, calculated using the energy released per gram of U-235.
277 Medium
277 of 380
Understanding +4 | -1
Assertion: The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometer.

Reason: Nuclear force is a short range force
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Option A is correct because the nuclear force is a short-range force, and its strength diminishes rapidly beyond a few femtometers.
278 Medium
278 of 380
Understanding +4 | -1
Mn{M_n} and Mp{M_p} represent mass of neutron and proton respectively. If an element having atomic mass MM has NN-neutrons and ZZ-protons, then the correct relation will be
View answer and explanation
Correct option: A
M<[N.Mn+Z.MP]M < \left[ {N.{M_n} + Z.{M_P}} \right]
Option A (M<[N.Mn+Z.MP]M < [N.{M_n} + Z.{M_P}]) is correct because the mass of the nucleus is less than the sum of the masses of its constituent nucleons due to the binding energy.
279 Medium
279 of 380
Application +4 | -1
If half life of radium is 77 days\mathrm{77 }\,\mathrm{days}. Its decay constant in day will be
View answer and explanation
Correct option: B
9  ×  10−3/day9\,\, \times \,\,{10^{ - 3}}/\mathrm{day}
Option B (9×10−3/day9 \times 10^{-3}/\mathrm{day}) is correct because it is calculated using the formula relating half-life and decay constant.
280 Medium
280 of 380
Application +4 | -1
The activity of a radioactive sample decreases to one-third of the original value in 9 years. What will be the activity after further 9 years ?
View answer and explanation
Correct option: C
A0/9\mathrm{A}_0 / 9
Option C (A0/9\mathrm{A}_0 / 9) is correct because after two half-lives (18 years), the activity will be reduced to one-ninth of the original value, given that it reduces to one-third in the first 9 years.
281 Medium
281 of 380
Application +4 | -1
Half-life of radioactive sample, when activity of material initially was 8 counts\mathrm{8 }\,\mathrm{counts} and after 3  h\mathrm{3}\,\,\mathrm{h} it becomes 1 counts\mathrm{1 }\,\mathrm{counts}, is
View answer and explanation
Correct option: B
1  h\mathrm{1}\,\,\mathrm{h}
Option B (1h\mathrm{1}\\ h) is correct because it is calculated by finding the number of half-lives that have elapsed and dividing the total time by that number.
282 Medium
282 of 380
Understanding +4 | -1
The ratio of mass densities of nuclei of 40Ca^{40}Ca and 16O^{16}O is close to
View answer and explanation
Correct option: D
1
Option D is correct because the mass density of nuclear matter is approximately constant for all nuclei.
283 Medium
283 of 380
Understanding +4 | -1
Which of the following is possible in a β+\beta^{+} decay?
View answer and explanation
Correct option: A
A proton may decay into neutron and a positron
Option A is correct because in β+\beta^+ decay, a proton transforms into a neutron, a positron, and a neutrino to conserve charge and baryon number.
284 Medium
284 of 380
Understanding +4 | -1
A nucleus decays by β+{\mathrm{\beta }^ + }-emission followed by a γ−emission\gamma - \mathrm{emission}. If the atomic and mass numbers of the parent nucleus are ZZ and AA respectively, the corresponding numbers for the daughter nucleus are respectively
View answer and explanation
Correct option: C
Z−1Z - 1 and AA
Option C (Z−1Z - 1 and AA) is correct because β+{\mathrm{\beta }^ + }-emission decreases the atomic number by 1, while γ\gamma-emission does not change either number.
285 Medium
285 of 380
Application +4 | -1
The fraction of the initial number of radioactive nuclei which remain undecayed after half of a half-life of the radioactive sample is
View answer and explanation
Correct option: A
12\frac{1}{{\sqrt 2 }}
Option A (12\frac{{1}}{{\sqrt 2 }}) is correct because after half of a half-life, the remaining fraction is calculated as N=N0/2t/T1/2=N0/2N = N_0 / 2^{t/T_{1/2}} = N_0 / \sqrt{2}.
286 Medium
286 of 380
Understanding +4 | -1
65Cu{ }^{65} {Cu} will turn into 66Cu{ }^{66} {Cu} if it is bombarded with
View answer and explanation
Correct option: B
Neutrons
Option B (Neutrons) is correct because bombarding 65Cu{ }^{65} {Cu} with neutrons adds a neutron to the nucleus, resulting in 66Cu{ }^{66} {Cu}.
287 Medium
287 of 380
Understanding +4 | -1
Mp{M_p} denotes the mass of a proton and Mn{M_n} that of a neutron. A given nucleus, of binding energy BB, contains ZZ protons and NN neutrons. The mass M(N,Z)M\left( {N,Z} \right) of the nucleus is given by (cc is the velocity of light)
View answer and explanation
Correct option: C
M(N,Z)=NMn+ZMp−B/c2M\left( {N,Z} \right) = N{M_n} + Z{M_p} - B/{c^2}
Option C is correct because the mass of a nucleus is less than the sum of the masses of its constituent nucleons due to the binding energy, which is released when the nucleus is formed.
288 Medium
288 of 380
Application +4 | -1
It is observed that only 6.25%6.25 \% of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample, in day −1{ }^{-1} ?
View answer and explanation
Correct option: A
0.173 day −10.173 \text { day }^{-1}
Option A (0.173 day −10.173 \text{{ day }}^{-1}) is correct because it is calculated using the half-life formula and the decay constant relationship.
289 Medium
289 of 380
Application +4 | -1
A radioactive substance contains 10000\mathrm{10000} nuclei and its half-life period is 20\mathrm{20} days. The number of nuclei present at the end of 10\mathrm{10} days is
View answer and explanation
Correct option: A
7070\mathrm{7070}
Option A (7070\mathrm{7070}) is correct because it is calculated using the radioactive decay formula N=N0(12)tTN = N_0 (\frac{1}{2})^{\frac{t}{T}}.
290 Medium
290 of 380
Analyzing +4 | -1
Consider the binding energy of 17Cl35{ }_{17} {Cl}^{35} and 15P31{ }_{15} {P}^{31} are 287.67 MeV and 262.48MeV262.48 {MeV}, respectively. Then,
View answer and explanation
Correct option: B
15P31{ }_{15} {P}^{31} is more stable than 17Cl35{ }_{17} {Cl}^{35}
Option B is correct because the stability of a nucleus is determined by the binding energy per nucleon, and 15P31{}_{15}P^{31} has a higher binding energy per nucleon than 17Cl35{}_{17}Cl^{35}.
291 Medium
291 of 380
Remembering +4 | -1
The sodium nucleus 11Na23{}_{11}N{a^{23}} contains
View answer and explanation
Correct option: D
12 neutrons
Option D (12 neutrons) is correct because the number of neutrons is calculated by subtracting the atomic number (number of protons) from the mass number.
292 Medium
292 of 380
Remembering +4 | -1
Nuclear forces are
View answer and explanation
Correct option: A
Short ranged attractive and charge independent
Option A (Short ranged attractive and charge independent) is correct because nuclear forces are strong, short-ranged, and act equally between protons and neutrons, making them charge-independent.
293 Medium
293 of 380
Application +4 | -1
Half – lives of two radioactive elements AA and BB are 20 min and 40 min, respectively. Initially, the samples have equal number of nuclei. After 80 min, the ratio of decayed numbers of AA and BB nuclei will be
View answer and explanation
Correct option: D
0.211111111
Option D (5:04) is correct because it represents the ratio of decayed nuclei after considering the different half-lives.
294 Medium
294 of 380
Remembering +4 | -1
Which of the following statements is true for nuclear forces?
View answer and explanation
Correct option: C
they are short range forces
Option C (they are short range forces) is correct because nuclear forces operate only over very short distances, approximately the size of the nucleus.
295 Medium
295 of 380
Understanding +4 | -1
Given below are two statements: one is labelled as
Assertion A and the other is labelled as Reason R

Assertion A : The binding energy per nucleon is
practically independent of the atomic number for
nuclei of mass number in the range 30 to 170.

Reason R : Nuclear force is short ranged.

In the light of the above statements, choose the
correct answer from the options given below
View answer and explanation
Correct option: B
Both A and R are true and R is the correct
explanation of A
Option B (Both A and R are true and R is the correct explanation of A) is correct because the short-range nature of the nuclear force explains the near-constant binding energy per nucleon.
296 Medium
296 of 380
Application +4 | -1
In half life of a radio isotope is 2  seconds\mathrm{2}\,\,\mathrm{seconds} and number of atoms are only 4\mathrm{4}, then after one half life remaining (without decay) atoms are probably
View answer and explanation
Correct option: B
2\mathrm{2}
Option B (2) is correct because after one half-life, half of the radioactive atoms will have decayed, leaving half of the original number.
297 Medium
297 of 380
Understanding +4 | -1
Assertion: Radioactive nuclei emit β−1\beta^{-1} particles.
Reason: Electrons exist inside the nucleus.
View answer and explanation
Correct option: C
If assertion is true but reason is false
Option C (If assertion is true but reason is false) is correct because radioactive nuclei do emit β−1\beta^{-1} particles, but electrons do not exist inside the nucleus.
298 Medium
298 of 380
Application +4 | -1
A count rate metre shows a count of 240\mathrm{240} per minute from a given radioactive source later the metre shows a count rate 0f 30  min⁡−130\,\,{\min ^{ - 1}}⁡ . The half-life of the source is
View answer and explanation
Correct option: C
20 min\mathrm{20}\,\mathrm{ min}
Option C (20 min) is correct because it is the half-life that satisfies the decay equation given the initial and final count rates over a period of 60 minutes.
299 Medium
299 of 380
Application +4 | -1
In a reactor, 2kg of 92U235_{92}{U^{235}} fuel is fully used up in 30 days. The energy released per fission is 200 MeV.
Given that, the Avogadro number, N=6.023×1026K−1 mol−1N = 6.023 \times {10^{26}}{K^{ - 1}}\,mo{l^{ - 1}} and 1  eV=1.6×10−19J.1\,\,eV = 1.6 \times {10^{ - 19}}J. The power output of the reactor is close to
View answer and explanation
Correct option: C
60 MW
Option C (60 MW) is correct because it is the closest value to the calculated power output of the reactor, considering the energy released per fission and the amount of fuel used.
300 Medium
300 of 380
Understanding +4 | -1
At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound cannot emit
View answer and explanation
Correct option: D
neutrons
Option D (neutrons) is correct because neutrons are neutral particles and are not deflected by magnetic fields.
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