Question

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

Answer 1

Hint: To solve this question we use basic formulas related to Nuclear Physics. First, we calculate the Momentum of emitted photons by using the formula and then after calculate the recoil energy of the nucleus with the help of previously calculated momentum. Some of them are mentioned below as we will discuss.

Formula used:
Energy = Planck's constant$ \times $wave frequency
(E = hf)
E = photon energy
$h$ = Planck's constant
$v$= wave frequency

Complete step-by-step solution:
As we know,
If the energy of light is given and the velocity of the photon is given then we calculate the Momentum of the emitted photon.
Now, we get:
Momentum of emitted photon = $\dfrac{{{\text{energy of light}}}}{{{\text{velocity of photon}}}}$= $\dfrac{{hv}}{c}$
Now, using the above two formula, we get
According to the momentum conservation value of the momentum of recoil, the nucleus should also be $\dfrac{{hv}}{c}$
but in the opposite direction to make net momentum zero before and after emission.
Now recoil energy = $\dfrac{1}{2}M{v^2}$=$\dfrac{1}{{2M}}{M^2}{v^2}$= $\dfrac{{{p^2}}}{{2M}}$
Put value of p in above equation-
recoil energy = $\dfrac{{{p^2}}}{{2M}}$
                        = ${\left( {\dfrac{{hv}}{c}} \right)^2}$$ \times \dfrac{1}{{2M}}$
                        = $\dfrac{{{h^2}{v^2}}}{{2M{c^2}}}$
Therefore, the recoil energy will be $\dfrac{{{h^2}{v^2}}}{{2M{c^2}}}$.
And if we observe the above equation then we notice that the recoil energy of the nucleus depends on the nucleus of mass M, the photon of frequency v.
Thus, option (C) is the correct answer.

Note: Recoil is the backward movement of a gun when it is discharged. In technical terms, the recoil momentum acquired by the gun exactly balances the forward momentum of the projectile and exhaust gases, according to Newton's third law, known as conservation of momentum. In hand-held small arms, the recoil momentum is transferred to the ground through the body of the shooter.