Question

If the work function of a metal is \(1.2\,eV\) and the stopping potential is \(1.8\,V\), find the frequency of incident light on the metal surface.

• A. \(4.84\times10^{14}\,Hz\)
• B. \(7.25\times10^{14}\,Hz\)
• C. \(1.45\times10^{14}\,Hz\)
• D. \(9.67\times10^{14}\,Hz\)

Hint:

For photoelectric effect problems, remember Einstein's equation: \[ \boxed{hf=\phi+eV_s} \] If the work function is given in electron volts, \[ \boxed{\text{Photon Energy (eV)}=\phi+V_s.} \] Then convert the energy into joules and use \[ \boxed{f=\frac{E}{h}} \] to calculate the frequency.

Answer 1

The Correct Option is B

Concept: The photoelectric effect is explained by Einstein's photoelectric equation, \[ hf=\phi+K_{\max}, \] where \[ h=\text{Planck's constant}, \] \[ f=\text{Frequency of incident radiation}, \] \[ \phi=\text{Work function of the metal}, \] \[ K_{\max}=\text{Maximum kinetic energy of the emitted electrons}. \] The maximum kinetic energy is related to the stopping potential by \[ K_{\max}=eV_s. \] If the work function is expressed in electron volts (eV), then \[ hf=\phi+V_s \quad (\text{in eV}). \] Also, \[ 1\,eV=1.6\times10^{-19}\,J, \] and \[ h=6.626\times10^{-34}\,Js. \]

Step 1: Write the given data.
Given, \[ \phi=1.2\,eV, \] \[ V_s=1.8\,V. \]

Step 2: Calculate the photon energy.
According to Einstein's equation, \[ E=\phi+eV_s. \] In electron volt, \[ E=1.2+1.8=3.0\,eV. \] Converting into joules, \[ E = 3\times1.6\times10^{-19} = 4.8\times10^{-19}\,J. \]

Step 3: Calculate the frequency of the incident light.
Using, \[ E=hf, \] we obtain \[ f = \frac{E}{h}. \] Substituting the values, \[ f = \frac{4.8\times10^{-19}} {6.626\times10^{-34}}. \] Therefore, \[ f = 7.24\times10^{14}\,Hz. \] Approximating, \[ \boxed{f=7.25\times10^{14}\,Hz.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (B)}}. \]