Suppose a $_{88}^{226} R a$ nucleus at rest and in ground state undergoes $α$ -decay to a $_{86}^{222} R n$ nucleus in its excited state. The kinetic energy of the emitted $α$ particle is found to be $4.44 M e V$ . $_{86}^{222} R n$ nucleus then goes to its ground state by $γ$ -decay. The energy of the emitted $γ$ photon is ____ keV..
[Given : atomic mass of $_{88}^{226} R a = 226.005 u$ , atomic mass of $_{88}^{222} R n = 222.000 u$ , atomic mass of $α$ particle = $4.000 u, 1 u = 931 M e V / c^{2}, c$ is speed of light]

Question

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Answer 1

Mass defect $Δ m = 226.005 - 222.000 - 4.000$
$= 0.005 a m u$
$∴ Q$ value = $0.005 \times 931.5 = 4.655 M e V$
Also $\frac{K . E _{α}}{K . E _{R n}} = \frac{m _{R n}}{m _{α}}$
$\Rightarrow K . E_{R n} = \frac{m _{α}}{m _{R n}} . K . E_{α} = \frac{4}{2 2 2} \times 4.44 = 0.08 M e V$
$∴$ Energy of $γ$ -Photon = $4.655 - (4.44 + 0.08)$
$= 0.135 M e V = 135 K e V$

Answer 2

Mass defect

Δ m = 226.005 - 222.000 - 4.000

= 0.005 a m u

∴ Q

value =

0.005 \times 931.5 = 4.655 M e V

Also

\frac{K . E _{α}}{K . E _{R n}} = \frac{m _{R n}}{m _{α}}

\Rightarrow K . E_{R n} = \frac{m _{α}}{m _{R n}} . K . E_{α} = \frac{4}{2 2 2} \times 4.44 = 0.08 M e V

∴

Energy of

γ

-Photon =

4.655 - (4.44 + 0.08)

= 0.135 M e V = 135 K e V