Question

In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is $2 \times {10^{ - 7}}\,{\text{m}}$ and stopping potential is $2.5\,{\text{V}}$. The threshold frequency of the metal in Hz. Approximately $\left( e \right) = 1.6 \times {10^{ - 19}}\,{\text{C}}$, $h = 6.6 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}$
A. $12 \times {10^{15}}$
B. $9 \times {10^{15}}$
C. $9 \times {10^{14}}$
D. $12 \times {10^{13}}$

Answer 1

Use the formula for total energy of the photon incident on the metallic plate in terms of work function and maximum kinetic energy of the photoelectron emitted from the metal surface. Also use the formula for work function and maximum kinetic energy of the photoelectron in terms of threshold frequency of metal and stopping potential respectively.

Formulae used:
The energy of the photon incident on a metallic surface for photoemission is
$\dfrac{{hc}}{\lambda } = \phi + K$ …… (1)
Here, $h$ is the Planck’s constant, $c$ is speed of light, $\lambda$ is wavelength of the incident photon, $\phi$ is work function of photoelectron and $K$ is maximum kinetic energy of the emitted photoelectron.
The work function $\phi$ of a photoelectron is
$\phi = h{\gamma _0}$ …… (2)
Here, $h$ is the Planck’s constant and ${\gamma _0}$ is the threshold frequency of the metal.
The maximum kinetic energy $K$ of the photoelectron emitted is
$K = e{V_0}$ …… (3)
Here, $e$ is the charge on the electron and ${V_0}$ is the stopping potential.

Complete step by step answer:
We have given that the wavelength of the incident light on the metallic surface is $2 \times {10^{ - 7}}\,{\text{m}}$ and the stopping potential is $2.5\,{\text{V}}$.
$\lambda = 2 \times {10^{ - 7}}\,{\text{m}}$, ${V_0} = 2.5\,{\text{V}}$
We have asked to determine the threshold frequency of the metal.
Substitute $h{\gamma _0}$ for $\phi$ and $e{V_0}$ for $K$ in equation (1).
$\dfrac{{hc}}{\lambda } = h{\gamma _0} + e{V_0}$

Rearrange the above for the threshold frequency of the metal.
$\Rightarrow {\gamma _0} = \dfrac{c}{\lambda } - \dfrac{{e{V_0}}}{h}$

Substitute $3 \times {10^8}\,{\text{m}}/s$ for $c$, $2 \times {10^{ - 7}}\,{\text{m}}$ for $\lambda$, $1.6 \times {10^{ - 19}}\,{\text{C}}$ for $e$, $2.5\,{\text{V}}$ for ${V_0}$ and $6.6 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}$ for $h$ in the above equation.
$\Rightarrow {\gamma _0} = \dfrac{{3 \times {{10}^8}\,{\text{m}}/s}}{{2 \times {{10}^{ - 7}}\,{\text{m}}}} - \dfrac{{\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)\left( {2.5\,{\text{V}}} \right)}}{{6.6 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}}}$
$\Rightarrow {\gamma _0} = 8.93 \times {10^{14}}\,{\text{Hz}}$
$\therefore {\gamma _0} \approx 9 \times {10^{14}}\,{\text{Hz}}$

Therefore, the threshold frequency of the metal is $9 \times {10^{14}}\,{\text{Hz}}$. Hence, the correct option is C.

Note: The students may try to determine the threshold frequency of the metal using the wavelength of incident light and speed of light. But the students should keep in mind that the frequency determined in this way is the frequency of the incident light wave and not the threshold frequency of the metal.