Explanation:
Expression for the kinetic energy is as follows.
K.E = [tex]\frac{1}{2}mv^{2}[/tex]
Now, total kinetic energy will be as follows.
K.E = [tex]2 \times \frac{1}{2}mv^{2}[/tex] = [tex]m \times v^{2}[/tex]
Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.
Relation between energy and photon is as follows.
Energy of photon = [tex]\frac{hc}{\lambda}[/tex]
= [tex]\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}[/tex]
= [tex]1.63 \times 10^{-18} J[/tex]
[tex]m \times v^{2} = 1.63 \times 10^{-18}[/tex]
v = [tex]\sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}[/tex]
= [tex]3.12 \times 10^{4}[/tex] m/s
Thus, we can conclude that atoms were moving at a speed of [tex]3.12 \times 10^{4}[/tex] m/s before the collision.
