Question

When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9V. If \[\dfrac{e}{m}\] for the electron is \[1.8 \times {10^{11}}Ck{g^{ - 1}}\], the maximum velocity of the ejected electron is:
A. \[6 \times {10^5}m{s^{ - 1}}\]
B. \[8 \times {10^5}m{s^{ - 1}}\]
C. \[1.8 \times {10^6}m{s^{ - 1}}\]
D. \[1.8 \times {10^5}m{s^{ - 1}}\]

Answer 1

Hint: When a photon strikes a metal, it has enough energy to overcome the attractive attraction that holds the valence electron to the shell of the metal's atom. In order to solve this problem, we are going to use the photoelectric effect equations.

Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[\phi \] is the work function of the metal.
\[{K_{\max }} = e{V_0}\]
Here e is the charge on the electron and \[{V_0}\] is the stopping potential.
\[K = \dfrac{{m{v^2}}}{2}\]
Here K is the kinetic energy of the body of mass m and moving with velocity v.

Complete step by step solution:
As the energy of the photon is inversely proportional to the wavelength, so for a minimum value of the energy, the wavelength should be the maximum allowed wavelength. The stopping potential is the potential which is able to stop the ejected electron. Using the work energy theorem, when a body is acted upon by external force then the change in kinetic energy of the body is equal to the work done on it.

The work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron. It is given as 9V. Let the mass of the electron is m and the velocity of the ejected electron is v.
\[K{E_{\max }} = e{V_0}\]
\[\Rightarrow \dfrac{{m{v^2}}}{2} = e{V_0}\]
\[\Rightarrow v = \sqrt {\dfrac{{2e{V_0}}}{m}} \]
\[\Rightarrow v = \sqrt {2{V_0}\left( {\dfrac{e}{m}} \right)} \]
\[\Rightarrow v = \sqrt {2 \times 9 \times 1.8 \times {{10}^{11}}}\,m{s^{ - 1}}\]
\[\therefore v = 1.8 \times {10^6}\,m{s^{ - 1}}\]
Hence, the velocity of the ejected electron is \[1.8 \times {10^6}\,m{s^{ - 1}}\].

Therefore, the correct option is C.

Note: The threshold frequency in a metal surface is the lowest frequency of incoming radiation necessary for the photoelectric effect to occur. There won't be any photoemission when the frequency of the incoming radiation is below the metal surface's threshold frequency.