Question

Stopping potential [V volts] is plotted against frequency of light used [v]. Find work function (eV). (h = 6.62 x 10^{-34} J x S)

• A. 0.212 eV
• B. 0.207 eV
• C. 0.414 eV
• D. 0.428 eV

Answer 1

Answer: (C)

0.414 eV

Answer 2

The relationship between stopping potential ($V_0$) and frequency ($\nu$) in the photoelectric effect is given by Einstein's photoelectric equation: $h\nu = \phi + eV_0$. Rearranging this, we get $V_0 = \left(\frac{h}{e}\right)\nu - \left(\frac{\phi}{e}\right)$. This is a linear equation. The x-intercept (where $V_0 = 0$) is the threshold frequency ($\nu_0$). From the graph, the line intersects the frequency axis at $\nu = 1 \times 10^{14}$ Hz. The work function ($\phi$) is given by $\phi = h\nu_0$. $\phi = (6.62 \times 10^{-34} \text{ J s}) \times (1 \times 10^{14} \text{ Hz}) = 6.62 \times 10^{-20}$ J. Converting to eV: $\phi (\text{eV}) = \frac{6.62 \times 10^{-20} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 0.413$ eV, which is approximately 0.414 eV.