Question:

The work functions of Aluminium and Gold are 4.1 eV and 5.1 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is 

Updated On: Mar 21, 2025
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The Correct Option is B

Solution and Explanation

Using \( K.E_{\text{max}} = eV_s = hf - \phi_0 \), where \(\phi_0\) is work function, \(V_s\) is stopping potential, and \(f\) is frequency, we get: \[ V_s = \frac{h}{e} f - \frac{\phi_0}{e}. \] Therefore, the slope \( m \) will be the same for all graphs and will be independent of \(\phi_0\).
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