Question

A free nucleus of mass 24 amu emits a gamma photon (When initially at rest). The energy of the photon is 7MeV. The recoil energy of the nucleus in keV is:
A. 2.2
B. 1.1
C. 3.1
D. 22

Answer 1

Hint: As we know that due to radioactive decay of some radioactive nucleus, a gamma-ray is emitted which contains a bunch of packets of electromagnetic energy called photons. Gamma rays have the smallest wavelength of electromagnetic waves and therefore they have the highest photon energy.

Complete step by step solution:
Here we have to express a hypothetical reaction of the nucleus when it moves from rest to excited state. Hence the reaction is:
$X \to {X^*} + \Upsilon $
Here X is the nucleus in state of rest, X* is the nucleus in an excited state and is the photon.
Hence, apply conservation of momentum,
${P_X} = {P_{{X^*}}} + {P_\Upsilon }$
$ \Rightarrow 0 = {P_{{X^*}}} + {P_\Upsilon }$ …… (I)
Here ${P_X}$ is the momentum of X, ${P_{{X^*}}}$ is the momentum of X* and ${P_\Upsilon }$ is the momentum of $\Upsilon $.
We know that the energy of the photon is given by
$E = Pc$ …… (II)
Here $E$ is the energy of the photon, $P$ is the momentum of the photon and $c$ is the speed of light.
We also know that the recoil energy of the nucleus is given by
\[\overline k = \overline {\dfrac{{{P^2}}}{{2\overline m }}} \] …… (III)
Here $\overline k $ is the recoil energy of the nucleus, $\overline P $ is the momentum of the nucleus and $\overline m $ is the mass of the nucleus.
We know that since X is at rest and hence its momentum is zero. So we can use the concept of equation (II) and equation (III) in equation (I) to obtain the final result that is that the modulus of the momentum of the nucleus is equal to the modulus of the momentum of the photon. So we can write it as,
\[
  \left| {{P_{{X^*}}}} \right| = \,\;\left| {{P_\Upsilon }} \right| \\
   \Rightarrow \sqrt {2km} = \dfrac{E}{c} \\
\]
\[ \Rightarrow k = \dfrac{{{E^2}}}{{2c{m^2}}}\] …… (IV)
Here $k$ is the recoil energy, $m$ is the mass of the nucleus, $E$ is the energy of the photon and $c$ is the speed of light.
Substitute $E=7MeV$, $c=3 \times 10^8 m/s$ and $m=24$ amu in equation (IV) to obtain the value of $k$.

$ k = \dfrac{{{E^2}}}{{2c{m^2}}} $
$ \Rightarrow k= \dfrac{(7Mev)^2}{2 \times 24 \times 1.67 \times 10^{-27} \times 3 \times 10^{8}}$
$\Rightarrow k=1.086 \times 10^{-3}MeV$
$\Rightarrow k=1.086 KeV$
$\Rightarrow k \approx 1.1KeV$

$\therefore$ The correct option is B.

Note:
The distance between two positions of the earth in the span of six months is equal to two astronomical units. We can also notice that over a course of the year some stars move a very small amount relative to other stars. The stars that are closer seem to move but the stars that are far away don’t seem to move. Actually all stars are moving through space but much more slowly than parallax so we don’t come to notice it.