Question

A free atom of iron emits ${K_\alpha }$ X-rays of energy $6.4\,{\text{KeV}}$. Calculate the recoil kinetic energy of the atom. Mass of the iron atom is $9.3 \times {10^{ - 26}}\,{\text{kg}}$ .

Answer 1

First of all, we will find out the momentum of the photon particle and then we will use the formula which relates kinetic energy and the momentum. By substituting the required values and manipulating accordingly, we can obtain the answer.

Complete step by step answer:
In the given problem, we are supplied with the following data:
The energy of the ${K_\alpha }$ X-rays is $6.4\,{\text{KeV}}$ .
Mass of the iron atom is $9.3 \times {10^{ - 26}}\,{\text{kg}}$ .
We are asked to find the recoil kinetic energy.
To begin with, we will first convert the energy given in electron volts to its equivalent joules.
We know,
$1\,{\text{eV}} = 1.6 \times {10^{ - 19}}\,{\text{J}}$
So,
6.4\,{\text{KeV}} = 6.4 \times {10^3} \times 1.6 \times {10^{ - 19}}\,{\text{J}} \\\ 6.4\,{\text{KeV}} = 10.24 \times {10^{ - 16}}\,{\text{J}} \\\
Since, the contains packets of electromagnetic energy (which contains photons) and these originate from the electron cloud present in an atom.
We will now calculate the momentum of a photon particle, which is given by the formula:
$p = \dfrac{E}{c}$ …… (1)
Where,
$p$ indicates momentum of a photon particle.
$E$ indicates energy of a photon particle.
$c$ indicates the velocity of the photon particle which is travelling at.
Now, we substitute the required values in the equation (1), we get:
p = \dfrac{E}{c} \\\ p = \dfrac{{10.24 \times {{10}^{ - 16}}\,{\text{J}}}}{{3 \times {{10}^8}\,{\text{m}}{{\text{s}}^{ - 1}}}} \\\ p = 3.41 \times {10^{ - 24}}\,{\text{kgm}}{{\text{s}}^{ - 1}} \\\
Therefore, the momentum of a photon particle is found to be $3.41 \times {10^{ - 24}}\,{\text{kgm}}{{\text{s}}^{ - 1}}$ .
We have, according to the principle of conservation of momentum, that momentum before impact is equal to the momentum after impact.
To find the recoil kinetic energy, we use the formula, which is given below:
$K.E = \dfrac{{{p^2}}}{{2m}}$ …… (2)
Where,
$K.E$ indicates the recoil kinetic energy of the iron atom.
$p$ indicates the momentum of the iron atom.
$m$ indicates the mass of an iron atom.
Now, we substitute the required values in equation (2) and we get:
K.E = \dfrac{{{p^2}}}{{2m}} \\\
\implies K.E = \dfrac{{{{\left( {3.41 \times {{10}^{ - 24}}\,{\text{kgm}}{{\text{s}}^{ - 1}}} \right)}^2}}}{{2 \times 9.3 \times {{10}^{ - 26}}\,{\text{kg}}}} \\\
\implies K.E = \dfrac{{1.16 \times {{10}^{ - 47}}}}{{2 \times 9.3 \times {{10}^{ - 26}}}}\,{\text{J}} \\\
\implies K.E = 6.23 \times {10^{ - 23}}\,{\text{J}} \\\
Therefore, the recoil kinetic energy of the atom of iron is found to be $6.23 \times {10^{ - 23}}\,{\text{J}}$ .
We will now convert joules into electron volts:
6.23 \times {10^{ - 23}}\,{\text{J}} \\\ = \dfrac{{6.23 \times {{10}^{ - 23}}\,{\text{J}}}}{{1.6 \times 10 - 19\,{\text{Je}}{{\text{V}}^{ - 1}}}} \\\ = 3.89 \times {10^{ - 4}}\,{\text{eV}} \\\ = 0.389 \times {10^{ - 3}}\,{\text{eV}} \\\
It can also be written as $0.389\,{\text{meV}}$ .
Hence, the recoil kinetic energy of the atom of iron is found to be $0.389\,{\text{meV}}$ .

Note:
While solving this problem, remember that the momentum associated with the photon particle is equal to the momentum associated with the atom of iron. The velocity associated with the photon particles is equal to the velocity of light, as light is an electromagnetic wave which contains photons.