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JEE Advanced 2013 Paper 2

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The mass of a nucleus

{}^A_Z

X is less than the sum of masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses

m_1

and

m_2

only if (

m_1 + m_2

) < M. Also two light nuclei of masses

m_3

and

m_4

can undergo complete fusion and form a heavy nucleus of mass M

'

only if (m

{}_3

+ m

{}_4

) > M

'

. The masses of some neutral atoms are given in the table below :

$\overset{1}{{}_1}$ H	1.007825u	$\overset{2}{{}_1}$ H	2.014102u	$\overset{3}{{}_1}$ H	3.016050u	$\overset{4}{{}_2}$ He	4.002603u
$\overset{6}{{}_3}$ Li	6.015123u	$\overset{7}{{}_3}$ Li	7.016004u	$\overset{70}{{}_30}$ Zn	69.925325u	$\overset{82}{{}_34}$ Se	81.916709u
$\overset{152}{{}_64}$ Gd	151.919803u	$\overset{206}{{}_84}$ Pb	205.974455u	$\overset{209}{{}_83}$ Bi	208.980388u	$\overset{210}{{}_84}$ Po	209.982876u

Section: Physics

Question : 15 of 60

Marks: +1, -0

The correct statement is :

[JEE Adv 2013 P2]

The nucleus $\overset{6}{_{3}}$ Li can emit an alpha partcle
The nucleus $\overset{210}{_{84}} \mathrm{P}_0$ can emit a proton
Deuteron and alpha particle can undergo complete fusion
The nuclei $\overset{70}{_{30}}$ Zn and $\overset{82}{_{34}}$ Se can undergo complete fusion

Validate

Solution:

(A)

^{7}_{3}\; \mathrm{Li} \;

→

_{2}^{4}\; \mathrm{He} \;

^{3}_{1}\; \mathrm{H} \;

Δm = [

M_{\mathrm{Li}} - M_{\mathrm{He}} - M_{\mathrm{H}^{3}}

]
= [6.01513 - 4.002603 - 3.016050]
= -1.003523u
Δm is negative so reaction is not possible.
(B)

_{84}^{210}\mathrm{Po}

→

_{83}^{209}\mathrm{Bi} + _{1}^{1}\mathrm{P}

Δm is negative so reaction is not possible.
(C)

_{1}^{2}\mathrm{H}

→

_{2}^{4}\mathrm{He} + _{3}^{6}\mathrm{Li}

Δm is positive so reaction is possible.
(D)

_{30}^{70}\mathrm{Zn}

_{34}^{82}\mathrm{Se} \rightarrow _{64}^{152}\mathrm{Gd}

Δm is positive so reaction is possible.

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