Skip to main content
Back to NUCLEI

Your Progress

0 / 351 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • NUCLEI

Nuclei practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 331-351 of 351 questions

Preparing practice and saved progress…

331 Medium
331 of 351
Application +4 | -1
In the reaction
7N14+α→8X17+1p1_7{\mathrm{N}^{14}} + \alpha { \to _8}{X^{17}}{ + _1}{p^1}
identify XX.
View answer and explanation
Correct option: A
O2{\mathrm{O}_2}
Option A (O2{\mathrm{O}_2}) is correct because balancing the nuclear reaction requires the product to be an isotope of Oxygen.
332 Medium
332 of 351
Remembering +4 | -1
Isotopes are atoms having
View answer and explanation
Correct option: A
Same number of protons but different number of neutrons
Option A (Same number of protons but different number of neutrons) is correct because isotopes have the same atomic number (protons) but different mass numbers (neutrons).
333 Medium
333 of 351
Application +4 | -1
If a star can convert all the He nuclei completely into oxygen nuclei, the energy released per oxygen nuclei is ( Mass of the nucleus is 4.0026    amu4.0026\;\,\,\mathrm{amu} and mass of oxygen nucleus is 15.9994    amu15.9994\;\,\,\mathrm{amu})
View answer and explanation
Correct option: C
10.24  MeV10.24\,\,MeV
Option C (10.24  MeV10.24 \,\,MeV) is correct because it is calculated from the mass defect when four Helium nuclei fuse to form one Oxygen nucleus, converted to energy using E=Δmc2E = \Delta m c^2.
334 Medium
334 of 351
Application +4 | -1
The half-life of Au198A{u^{198}} is 2.7 days. The activity of 1.50 mg of Au198A{u^{198}} if its atomic weight is 198 g mol−1198\,g\,mo{l^{ - 1}} is (NA = 6×1023 / mol)\left( {{N_A}\, = \,6 \times {{10}^{23}}\,/\,mol} \right)
View answer and explanation
Correct option: B
357 CiCi
Option B (357 Ci) is correct because the activity is calculated using the half-life, amount, and atomic weight of Au198.
335 Medium
335 of 351
Understanding +4 | -1
The nucleus of atomic mass AA and atomic number ZZ emits a β\mathrm{\beta }-particle. The atomic mass and atomic number of the resulting nucleus are
View answer and explanation
Correct option: C
A,Z+1A,Z + 1
Option C is correct because in beta decay, the mass number (A) remains the same, while the atomic number (Z) increases by 1.
336 Medium
336 of 351
Understanding +4 | -1
The volume of the nucleus having a mass number A\mathrm{A} is directly proportional to
View answer and explanation
Correct option: C
AA
Option C (AA) is correct because the volume of a nucleus is directly proportional to its mass number A.
337 Medium
337 of 351
Understanding +4 | -1
Consider two nuclei of the same radioactive nuclide. One of the nuclei was created in a supernova explosion 5   billion  years\mathrm{5}\,\,\,\mathrm{billion}\,\,\mathrm{years} ago. The probability of decay during the next time is
View answer and explanation
Correct option: D
Independent of the time of creation
The correct answer is D (Independent of the time of creation) because radioactive decay is a random process and depends only on the decay constant.
338 Medium
338 of 351
Understanding +4 | -1
The fission of 235U^{235}U can be triggered by the absorption of slow neutrons by a nucleus. Similarly a slow proton can also be used. This statement is
View answer and explanation
Correct option: B
Wrong
Wrong is correct because fission requires a neutron to initiate the reaction, not a proton due to electrostatic repulsion.
339 Medium
339 of 351
Understanding +4 | -1
In 88Ra226_{88}R{a^{226}} nucleus, there are
View answer and explanation
Correct option: B
138\mathrm{138} neutrons and 88\mathrm{88} protons
Option B is correct because the number of protons is equal to the atomic number (88), and the number of neutrons is the mass number (226) minus the atomic number (88), which equals 138.
340 Medium
340 of 351
Understanding +4 | -1
According to the quark model, it is possible to build all the hadrons using
View answer and explanation
Correct option: C
3 quarks\mathrm{3 }\,\mathrm{quarks} and 3 antiquarks\mathrm{3 }\,\mathrm{antiquarks}
Option C (If Assertion is True but the Reason is False) is correct because the assertion about standing waves is correct, but the reason is false as amplitude does depend on position.
341 Medium
341 of 351
Application +4 | -1
Two radioactive sources AA and BB of half lives 1  h  and  2  h\mathrm{1}\,\,\mathrm{h }\,\,\mathrm{and}\,\,\mathrm{ 2}\,\,\mathrm{h} respectively initially contain the same number of radioactive atoms. At the end of two hours, their rates of disintegration are in the ratio of
View answer and explanation
Correct option: C
1:21:2
Option C (1:21:2) is correct because the ratio of disintegration rates depends on the number of radioactive atoms remaining after two hours, considering their respective half-lives.
342 Medium
342 of 351
Remembering +4 | -1
Which of the following isotopes is normally fissionable
View answer and explanation
Correct option: C
92U235_{92}{U^{235}}
Option C (92U235_{92}{U^{235}}) is correct because Uranium-235 is the only naturally occurring isotope that is readily fissionable by slow neutrons.
343 Medium
343 of 351
Application +4 | -1
The energy released in the fission of 1 Kg\mathrm{1}\,\mathrm{Kg} of _92 92U235_{92}{\mathrm{U}^{235}} is (energy per fission =200 MeV\mathrm{ = 200 }\,\mathrm{MeV})
View answer and explanation
Correct option: C
8.2  ×  1013  J8.2\,\, \times \,\,{10^{13}}\,\,\mathrm{J}
Option C (8.2×1013J8.2 \times 10^{13} J) is correct because it accurately calculates the energy released from 1 kg of Uranium-235 using the energy per fission and Avogadro's number.
344 Medium
344 of 351
Understanding +4 | -1
The gravitational force between a H{H}-atom and another particle of mass mm will be given by Newton's law : F=GM⋅mr2F=G \frac{M \cdot m}{r^2}, where rr is (in km{km} ) and
View answer and explanation
Correct option: B
M=mproton +melectron −Bc2(B=13.6eV)M=m_{\text {proton }}+m_{\text {electron }}-\frac{B}{c^2}(B=13.6 {eV})
Option B (M=mproton +melectron −Bc2(B=13.6eV)M=m_{\text {proton }}+m_{\text {electron }}-\frac{B}{c^2}(B=13.6 {eV})) is correct because the mass of the hydrogen atom is slightly less than the sum of the masses of its constituents due to binding energy.
345 Medium
345 of 351
Understanding +4 | -1
The difference between U235  and  U238{\mathrm{U}^{235}}\,\,\mathrm{and}\,\,{\mathrm{U}^{238}} atom is that
View answer and explanation
Correct option: D
U238{\mathrm{U}^{238}} contains 3\mathrm{3} more neutrons
Option D (U238{\mathrm{U}^{238}} contains 3\mathrm{3} more neutrons) is correct because isotopes of an element differ in the number of neutrons.
346 Medium
346 of 351
Understanding +4 | -1
The half-life of 131I{ }^{131}I is 8 days. Given a sample of 131I{ }^{131} {I} at time t=0t=0, we can assert that
View answer and explanation
Correct option: D
A given nucleus may decay at any time after t=0t = 0
Option D is correct because radioactive decay is a probabilistic process, and any given nucleus can decay at any time after t=0.
347 Medium
347 of 351
Understanding +4 | -1
Assertion: In pair annihilation, an electron and position destroy each other and produce 2γ2 \gamma - rays.

Reason: 2γ2 \gamma-rays produced will have equal momentum in opposite direction and momentum is conserved.
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 the annihilation process and momentum conservation are true, but the reason doesn't directly explain the annihilation.
348 Medium
348 of 351
Application +4 | -1
For radioactive material, the half-life is 10 minutes. If initially, there are 600 number of nuclei, the me taken (in minutes) for the disintegration of 450 nuclei is :
View answer and explanation
Correct option: A
20
Option A (20) is correct because after 2 half-lives, 450 nuclei will have disintegrated from an initial 600.
349 Medium
349 of 351
Application +4 | -1
Find the maximum energy that a beta particle can have in the following decay

176Lu→176Hf+e+vˉ. { }^{176} {Lu} \rightarrow{ }^{176} {Hf}+e+\bar{v} \text {. }

Atomic mass of 176Lu{ }^{176} {Lu} is 175.942694u175.942694 \mathrm{u} and that of 176Hf{ }^{176} {Hf} is 175.941420u175.941420 \mathrm{u}.
View answer and explanation
Correct option: A
1.182 MeV1 . 182 \mathrm{~MeV}
Option A (1.182 MeV1 . 182 \mathrm{~MeV}) is correct because it is calculated by finding the mass difference between the parent and daughter nuclei and converting it to energy using E=mc2E=mc^2.
350 Medium
350 of 351
Understanding +4 | -1
After five half lives what will be the fraction of initial substance
View answer and explanation
Correct option: B
(12)5{\left( {\frac{1}{2}} \right)^5}
Option B is correct because after each half-life, the amount of substance remaining is halved. After five half-lives, the fraction remaining is (1/2)5.
351 Medium
351 of 351
Understanding +4 | -1
The force acting between proton and proton inside the nucleus is
View answer and explanation
Correct option: C
Both
Option C is correct because both the Coulombic (electrostatic) force and the nuclear force act between protons inside the nucleus.
Revision notes and key takeaways

Review NUCLEI in your textbook or class notes, then use this set to check your understanding.

Showing 331-351 of 351 questions

Your practice

0 / 351 answered · 351 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.