*Simulator***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]**
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) $\mathrm{c} \sqrt{\frac{\Delta \mathrm{m}}{\mathrm{M}}}$

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

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**Sol.**(4) Because energy is releasing $\Rightarrow$ Binding energy per nucleon of product> that of parent $\Rightarrow \mathrm{E}_{2}>$ $\mathrm{E}_{1}$

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]**
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]**
Q.

**Statement-1:**A nucleus having energy $\mathrm{E}_{1}$ decays be $\beta^{-}$ emissionto 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]**
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

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**Sol.**(2) Released energy $=\left[1.6747 \times 10^{-27}-1.6725 \times 10^{-27}-9 \times 10^{-31}\right]$ $\times\left(3 \times 10^{8}\right)^{2} \mathrm{J}=0.73 \mathrm{MeV}$

Give atleast proper solution.

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