Question 196: Atoms & Nuclei

Question

A nucleus of mass M emits $$\gamma$$ -ray photon of frequency 'v'. The loss of internal energy by the nucleus is :
[Take 'c' as the speed of electromagnetic wave]

Accepted Answer

Energy of $$\gamma$$-ray, E
$$\gamma$$
= hv
and momentum of $$\gamma$$-ray, $${p_\gamma } = {h \over \lambda }$$ .... (i)
As, $${p_\gamma } = {{{E_\gamma }} \over c} = {{hv} \over c}$$ ..... (ii)
From Eqs. (i) and (ii), we get
$${p_\gamma } = {{hv} \over c} = {h \over \lambda }$$ [$$\because$$ $$\lambda  = {c \over v}$$]
Since, during the emission of $$\gamma$$-ray photon, momentum is conserved.
$$	herefore$$ p
$$\gamma$$
+ p
decayed nuclei
= 0
$$\Rightarrow$$ p
$$\gamma$$
= p
decayed nuclei
$$ \Rightarrow {{hv} \over c}$$ = p
decayed nuclei
..... (iii)
Kinetic energy of decayed nuclei,
$$KE = {1 \over 2}M{v^2} = {{(p_{decayed\,nuclei}^2)} \over {2M}}$$ ..... (iv)
From Eqs. (iii) and (iv), we get
$$ \Rightarrow KE = {1 \over {2M}}{\left[ {{{hv} \over c}} ight]^2}$$
$$	herefore$$ Loss in internal energy = E
$$\gamma$$
+ KE
decayed nuclei
$$ = hv + {1 \over {2M}}{\left[ {{{hv} \over c}} ight]^2} = hv\left[ {1 + {{hv} \over {2M{c^2}}}} ight]$$

Answer

hv

Answer

$$hv\left[ {1 + {{hv} \over {2M{c^2}}}} \right]$$

Answer

$$hv\left[ {1 - {{hv} \over {2M{c^2}}}} \right]$$

Answer

0

Atoms & Nuclei

A nucleus of mass M emits $\gamma$ -ray photon of frequency 'v'. The loss of internal energy by the nucleus is : [Take 'c' as the speed of electromagnetic wave]

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