Nuclear Physics – JEE Main Previous Year Questions with Solutions

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Previous Years AIEEE/JEE Mains Questions

Q. The above is a plot of binding energy per nucleon $\mathrm{E}_{\mathrm{b}}$, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions :

(i) $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\varepsilon$

(ii) $\mathrm{C} \rightarrow \mathrm{A}+\mathrm{B}+\varepsilon$

(iii) $\mathrm{D}+\mathrm{E} \rightarrow \mathrm{F}+\varepsilon$

$(\mathrm{iv}) \mathrm{F} \rightarrow \mathrm{D}+\mathrm{E}+\varepsilon$

where $\varepsilon$ is the energy released ? In which reactions is $\varepsilon$ positive ?

(1) (ii) and (iv) (2) (ii) and (iii) (3) (i) and (iv) (4) (i) and (iii)

[AIEEE – 2009]

Sol. (3)

Q. The speed of daughter nuclei is :-

(1) $\mathrm{c} \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}+\Delta \mathrm{m}}}$

(2) $\mathrm{c} \frac{\Delta \mathrm{m}}{\mathrm{M}+\Delta \mathrm{m}}$

(3) $\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}$

(4) $\mathrm{c} \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}}}$


Sol. (3)

Total kinetic energy of products

$=$ Total energy released $\frac{\mathrm{p}^{2}}{2 \mathrm{m}}+\frac{\mathrm{p}^{2}}{2 \mathrm{m}}$

$\left.=(\text { mass defect }) \mathrm{c}^{2} \text { (where } \mathrm{m}=\frac{\mathrm{M}}{2} \text { given }\right)$

$\Rightarrow 2\left(\frac{\mathrm{p}^{2}}{2 \mathrm{m}}\right)=\left[(\mathrm{M}+\Delta \mathrm{m})-\left(\frac{\mathrm{M}}{2}+\frac{\mathrm{M}}{2}\right)\right] \times \mathrm{c}^{2}$

$\Rightarrow 2 \times\left[\frac{\mathrm{p}^{2}}{2\left(\frac{\mathrm{M}}{2}\right)}\right]=(\Delta \mathrm{m}) \mathrm{c}^{2}$

$\Rightarrow \frac{2\left(\frac{\mathrm{M}}{2} \mathrm{v}\right)^{2}}{\mathrm{M}}=(\Delta \mathrm{m}) \mathrm{c}^{2} \Rightarrow \mathrm{v}=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}$

Q. The binding energy per nucleon for the parent nucleus is $E_{1}$ an that for the daughter nuclei is $\mathrm{E}_{2}$. Then:-

(1) $\mathrm{E}_{1}=2 \mathrm{E}_{2}$

$(2) \mathrm{E}_{2}=2 \mathrm{E}_{1}$

(3) $\mathrm{E}_{1}>\mathrm{E}_{2}$

$(4) \mathrm{E}_{2}>\mathrm{E}_{1}$

[AIEEE – 2010]

Sol. (4)

Because energy is releasing $\Rightarrow$ Binding energy per nucleon of product> that of parent $\Rightarrow \mathrm{E}_{2}>$


Q. A radioactive nucleus (initial mass number A and atomic number Z) emits 3 -particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be:-

(1) $\frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-2}$

(2) $\frac{\mathrm{A}-\mathrm{Z}-8}{\mathrm{Z}-4}$

(3) $\frac{\mathrm{A}-\mathrm{Z}-4}{\mathrm{Z}-8}$

(4) $\frac{\mathrm{A}-\mathrm{Z}-12}{\mathrm{Z}-4}$

[AIEEE – 2010]

Sol. (3)

Q. After absorbing a slowly moving neutron of mass $\left.\mathrm{m}_{\mathrm{N}} \text { (momentum } \sim 0\right)$ a nucleus of mass M breaks into two nuclei of masses $\mathrm{m}_{1}$ and $5 \mathrm{m}_{1}\left(6 \mathrm{m}_{1}=\mathrm{M}+\mathrm{m}_{\mathrm{N}}\right)$, respectively. If the de Broglie wavelength of the nucleus with mass $\mathrm{m}_{1}$ is $\lambda$, then de Broglie wavelength of the other nucleus will be:-

(1) $25 \lambda$

(2) $5 \lambda$

(3) $\frac{\lambda}{5}$

( 4)$\lambda$

[AIEEE – 2011]

Sol. (4)

Q. Statement-1: A nucleus having energy $\mathrm{E}_{1}$ decays be $\beta^{-}$ emissionto daughter nucleus having energy $E_{2}$, but the – rays are emitted with a continuous energy spectrum having end point energy $\mathrm{E}_{1}-\mathrm{E}_{2}$.

Statement-1: To conserve energy and momentum in -decay at least three particles must take part in the transformation.

(1) Statement-1 is incorrect, statement-2 is correct

(2) Statement-1 is correct, statement-2 is incorrect

(3) Statement-1 is correct, statement-2 correct; statement-2 is the correct explanation of statement-1

(4) Statement-1 is correct, statement-2 is correct; statement -2 is not the correct explanation of statement-1.

[AIEEE – 2011]

Sol. (3)

Q. Assume that a neutron breaks into a proton and an electron. The energy released during this process is :

(Mass of neutron $=1.6725 \times 10^{-27} \mathrm{kg}$

Mass of proton $=1.6725 \times 10^{-27} \mathrm{kg}$

Mass of electron $\left.=9 \times 10^{-31} \mathrm{kg}\right)$

(1) 5.4 MeV (2) 0.73 MeV (3) 7.10 MeV (4) 6.30 MeV

[AIEEE – 2012]

Sol. (2)

Released energy

$=\left[1.6747 \times 10^{-27}-1.6725 \times 10^{-27}-9 \times 10^{-31}\right]$

$\times\left(3 \times 10^{8}\right)^{2} \mathrm{J}=0.73 \mathrm{MeV}$


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  • June 27, 2020 at 5:08 pm

    please post question upto 2020

  • June 13, 2020 at 11:27 pm

    Please post the latest questions

  • May 3, 2020 at 7:30 am

    Is it helpful