Question

Based on given graph between stopping potential and frequency of irradiation, work function of metal is equal to

• A. 1 eV
• B. 3 eV
• C. 2 eV
• D. 4 eV

Answer 1

The Correct Option is C

To determine the work function of the metal from the given graph of stopping potential (V_s) versus frequency (ν), we can use the photoelectric equation:

\(V_s = \frac{h}{e} \nu - \frac{\phi}{e}\)

Where:

• \(V_s\) is the stopping potential,
• \(h\) is Planck's constant,
• \(\nu\) is the frequency of the incident light,
• \(\phi\) is the work function of the metal,
• \(e\) is the charge of an electron.

By examining the graph, we observe that the intercept on the frequency axis is approximately \(5 \times 10^{14} \, \text{Hz}\). This frequency is called the threshold frequency (ν₀).

The work function (\(\phi\)) is related to the threshold frequency by the equation:

\(\phi = h \nu_0\)

Since the threshold frequency ν₀ is the frequency at which the stopping potential becomes zero, this gives us:

Substituting the values:

\(\phi = (6.626 \times 10^{-34} \, \text{Js}) \times (5 \times 10^{14} \, \text{Hz})\)

Converting to electron volts, knowing \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\):

\(\phi = \frac{(3.313 \times 10^{-19} \, \text{J})}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 2 \, \text{eV}\)

Therefore, the work function of the metal is 2 eV.