Question

The photoelectric threshold wavelength of silver is $$3250\times {10}^{-10}m$$. The velocity of the electron ejected from a silver surface by the ultraviolet light of wavelength $$2536\times {10}^{-10}m$$is:
(Given $$h=4.14\times {10}^{-15}eVs$$ and $$c=3\times {10}^8\text{ m se}{\text{c}}^{-1}$$)
A. $$\approx 0.3\times {10}^{6}m{s}^{-1}$$
B. $$\approx 6\times {10}^{5}m{s}^{-1}$$
C. $$\approx 6\times {10}^{6}m{s}^{-1}$$
D. $$\approx 61\times {10}^{3}m{s}^{-1}$$

Answer 1

Hint: Threshold frequency is the minimum frequency at which radiation produces a photoelectric effect. In the photoelectric effect, the light is incident on the metal surface to eject the electrons depending upon the frequency. An increase in the frequency of the incident light increases the kinetic energy of the emitted electrons; hence a greater retarding potential is required to stop the emission.
According to the Einstein theory, when a photon falls on the surface of a metal plate, the entire photon’s energy is transferred to the electron. This can be written as
The energy of Photon = Energy required to eject electron + Maximum kinetic energy
$$\begin{gathered} E=W+KE \\ h\nu =W+KE \\ KE=h\nu -KE \end{gathered}$$
The energy carried out by each particle of light is dependent on the light’s frequency ($$\nu$$) is $$E=h\nu$$
The energy of a photon with frequency$${\nu }_0$$ must be the work function $$w=h{\nu }_0$$
Hence the maximum kinetic energy equation becomes $$KE={\displaystyle \frac{1}{2}}m{v}^2=h\nu -h{\nu }_0$$

Complete step by step answer:
Given that the wavelengths of the silver and the ultraviolet light are:
$$\lambda =2536\times {10}^{-10}m=2536{\mathrm{A}}^{\circ }$$
$${\lambda }_0=3250\times {10}^{-10}m=3250{\mathrm{A}}^{\circ }$$
Using maximum kinetic energy equation
$$\begin{gathered} {\displaystyle \frac{1}{2}}m{v}^2=h\nu -h{\nu }_0 \\ =hc\left[{\displaystyle \frac{1}{\lambda }}-{\displaystyle \frac{1}{{\lambda }_0}}\right]\left[As,v={\displaystyle \frac{c}{\lambda }}\right] \end{gathered}$$
This can be re-written as:
$$\begin{gathered} v^2={\displaystyle \frac{2hc}{m}}\left[{\displaystyle \frac{1}{\lambda }}-{\displaystyle \frac{1}{{\lambda }_0}}\right] \\ v=\sqrt{{\displaystyle \frac{2hc}{m}}\left[{\displaystyle \frac{1}{\lambda }}-{\displaystyle \frac{1}{{\lambda }_0}}\right]} \\ =\sqrt{{\displaystyle \frac{2\times \left(4.14\times {10}^{-15}\times 1.6\times {10}^{-19}\right)\times 3\times {10}^8}{9.1\times {10}^{-31}}}\left[{\displaystyle \frac{1}{2536\times {10}^{-10}}}-{\displaystyle \frac{1}{3250\times {10}^{-10}}}\right]} \\ =\sqrt{{\displaystyle \frac{3.974\times {10}^{-25}}{9.1\times {10}^{-31}}}\left[{\displaystyle \frac{1}{2536\times {10}^{-10}}}-{\displaystyle \frac{1}{3250\times {10}^{-10}}}\right]} \\ =\sqrt{2.72\times {10}^{24}\left[{\displaystyle \frac{714\times {10}^{-10}}{8242000\times {10}^{-20}}}\right]} \\ =\sqrt{4.3\times {10}^5\times 8.66\times {10}^5} \\ =\sqrt{37.24\times {10}^{10}} \\ =6.1\times {10}^5\text{m se}{\text{c}}^{-1} \end{gathered}$$
Hence the velocity of the electron ejected from a silver surface $$6.1\times {10}^5\text{m se}{\text{c}}^{-1}$$

So, the correct answer is “Option B”.

Note:
Students must note that the threshold frequency depends on the nature of the material used of the emitted electron. If the frequency of the light is less than the threshold frequency, then even an increase in intensity does not cause the emission of an electron.