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The work function of potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 Å photo electrons are emitted. The stopping potential of photo electrons is [Plank’s constant h = 6.6 × 10^–34 Js,
1 eV=1.6 × 10^–19 J, C = 3 ×10⁸ ms^–1]

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0.75 V

1.75 V

2.5 V

3.75 V

answer is B.

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Detailed Solution

$Stopping potential = (\frac{hc}{λ} - ϕ) / e$

$= (\frac{1240}{330} - 2) / e (hc = 1240 eV - nm)$

$\begin{aligned} = 1.75 eV / e \\ = 1.75 V \end{aligned}$

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