Question

The variation of stopping potential (V0) as a function of the frequency (v) of the incident light for a metal is shown in figure. The work function of the surface is

• A. 1.36 eV
• B. 2.98 eV
• C. 2.07 eV
• D. 18.6 eV

Answer 1

The Correct Option is C

Step 1: Understanding the photoelectric equation.
In the photoelectric effect, the stopping potential (\( V_0 \)) is related to the frequency (\( v \)) of the incident light by the equation: \[ eV_0 = h(v - v_{\text{th}}) \] Where:
\( V_0 \) is the stopping potential (in volts).
\( v \) is the frequency of the incident light (in Hz).
\( v_{\text{th}} \) is the threshold frequency (below which no photoelectric emission occurs).
\( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)).
\( h \) is Planck's constant (\( 6.6 \times 10^{-34} \, \text{J} \cdot \text{s} \)).
Step 2: Identifying the threshold frequency.
From the graph, we can observe that the stopping potential becomes non-zero at a frequency of approximately \( 5 \times 10^{14} \, \text{Hz} \). This is the threshold frequency \( v_{\text{th}} \).
Step 3: Calculating the work function.
At the threshold frequency, the stopping potential is zero. We use the equation: \[ \phi = h v_{\text{th}} \] Substituting the values: \[ \phi = (6.6 \times 10^{-34}) \times (5 \times 10^{14}) = 33 \times 10^{-20} \, \text{J} \] \[ \phi = 3.3 \times 10^{-19} \, \text{J} \] To convert this to eV, divide by the charge of the electron: \[ \phi = \frac{3.3 \times 10^{-19}}{1.6 \times 10^{-19}} \, \text{eV} = 2.07 \, \text{eV} \] Thus, the work function is \( \phi = 2.07 \, \text{eV} \).