The kinetic energy (in keV) of the alpha particle, when the nucleus at rest undergoes alpha decay, is [2013]

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 $m_{1}$ 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:

Q. The kinetic energy (in keV) of the alpha particle, when the nucleus ${}_{84}^{210}{Po}$ at rest undergoes alpha decay, is [2013]

Smart Achievers · Accepted Answer

(1)

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

$Mass defect$

$Δ m = [209.982876 - (205.974455 + 4.002603)] = 0.005818 μ$

$= 5.422 MeV = 5422 keV$

$∴$ $Q = (Δ m) \times 932 = (0.005818) \times 932$

$K_{α} = (\frac{A - 4}{A}) Q = (\frac{210 - 4}{210}) \times 5422$

$∴$ $K_{α} = 5319 keV$

Solution

1 5319

2 5422

3 5707

4 5818

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