Question

The mass of a nucleus $_{Z}^{A} X$ is less than the sum of the 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 m₁ and m₂ only if (m₁ + m₂) < M. Also, two light nuclei of masses m₃ and m₄ can undergo complete fusion and form a heavy nucleus of mass M’ only if (m₃ + m₄) > M'. The masses of some neutral atoms are given in the table below :

¹₁H	1.007825u	²₁H	2.014102u	³₁H	3.016050u	⁴₂He	4.002603u
⁶₃Li	6.015123u	⁷₃Li	7.016004u	⁷⁰₃₀Zn	69.925325u	⁸²₃₄Se	81.916709u
¹⁵²₆₄Gd	151.919803u	²⁰⁶₈₂Pb	205.974455u	²⁰⁹₈₃Bi	208.980388u	²¹⁰₈₄Po	209.982876u

(1u = 932 MeV/c²)

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Linked Question 1

The correct statement is

The nuclei $_{30}^{70} Zn and_{34}^{82} Se$ can undergo complete fusion.

The nucleus $_{3}^{6} Li$ can emit an alpha particle.

Deuteron and alpha particle can undergo complete fusion.

The nucleus $_{84}^{210} Po$ can emit a proton.

$_{3}^{6} Li \to_{1}^{2} H +_{2}^{4} He m_{daughter} {>m}_{parent}$

$_{84}^{210} Po \to_{83}^{209} Bi +_{1}^{1} p m_{daughter >} m_{parent}$

$_{1}^{2} H +_{2}^{4} He \to_{3}^{6} Hi m_{gauthter} < m_{parent}$

$_{30}^{70} Zn +_{34}^{82} Se \to_{64}^{152} Gd m_{daughter >} m_{parent}$

Linked Question 2

The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210} Po$ at rest undergoes alpha decay, is

5818

5319

5707

5422

$_{84}^{210} Po \to_{82}^{206} Pb +_{2}^{4} He$

Q = (m_Po – m_pb – m_$α$) c²

= 209.982876

205.974455

4.008421

– 4.002603

= 0.005818 × 932 MeV

= 5422 keV

Almost 100 keV go to nucleus

$\Rightarrow k_{α}$ = 5319 keV