Answer :
Answer:
The velocity of the photo electron is [tex]6.11\times 10^5\ m/s[/tex].
Explanation:
Given that,
Supplied energy, [tex]E_s=5.3\times 10^{-19}\ J[/tex]
Minimum energy of the electron to escape from the metal, [tex]E_e=3.6\times 10^{-19}\ J[/tex]
We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,
[tex]5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J[/tex]
The formula of kinetic energy is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s[/tex]
So, the velocity of the photo electron is [tex]6.11\times 10^5\ m/s[/tex].