Question

A radioactive nucleus of mass M emits a photon of frequency $\nu $ and the nucleus recoils. The recoil energy will be
${\text{A}}{\text{. h}}\nu $
${\text{B}}{\text{. M}}{{\text{c}}^{\text{2}}} - h\nu $
${\text{C}}{\text{. }}\dfrac{{{h^2}{\nu ^2}}}{{2M{c^2}}}$
${\text{D}}{\text{. }}$Zero

Answer 1

Hint: The energy of emitted photon is given by the formula, $E = h\nu $ and for a photon its momentum is given as Energy/velocity of light which is nothing but $\dfrac{E}{c}$ .Then substitute the value of E from above equation and then momentum of photon will be given by, $p = \dfrac{{h\nu }}{c}$ and recoil energy is given by the formula, ${E_{rec}} = \dfrac{{{p^2}}}{{2M}}$ .Use this concept and solve the question.

Complete step by step answer:
 $E = h\nu $ , $p = \dfrac{{h\nu }}{c}$ , ${E_{rec}} = \dfrac{{{p^2}}}{{2M}}$
We have been given that a radioactive nucleus of mass M emits a photon of frequency $\nu $ and the nucleus recoils.
To find – Recoil energy
Now, we know that the energy of photon is given by the $E = h\nu $
Also, the momentum of a photon is $p = \dfrac{E}{c}$
Here, p is the momentum of the photon, E is the energy and c is the velocity of light.
Now, substituting the value of E from above equation, we get momentum as-
$p = \dfrac{{h\nu }}{c}$
Here, h is the Planck’s constant, $\nu $ is the frequency of emission of photons.

Now, according to the conservation of momentum, value of the momentum of the recoil nucleus should also be $p = \dfrac{{h\nu }}{c}$ but in the opposite direction to make the net momentum zero before and after emission.
Now, the recoil energy is given by ${E_{rec}} = \dfrac{{{p^2}}}{{2M}}$ where p is the recoil momentum, M is the mass.
So, putting $p = \dfrac{{h\nu }}{c}$ in the formula of recoil energy we get-
${E_{rec}} = \dfrac{{{{\left( {\dfrac{{h\nu }}{c}} \right)}^2}}}{{2M}} = \dfrac{{{h^2}{\nu ^2}}}{{2M{c^2}}}$
So, the correct answer is “Option C”.

Note: Whenever such types of questions appear, then always first write down the things and the information given in the question and then as mentioned in the solution, use the formula of momentum of photon which will be equal to the recoil momentum, and put it in the formula of recoil energy i.e., ${E_{rec}} = \dfrac{{{p^2}}}{{2M}}$ . Also, the recoil momentum is equal and opposite to the momentum of the photon due to the conservation of momentum.