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^–34Js, 1eV = 1.6 × 10^–19J, 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

From Einstein's photoelectric equation, the stopping potential is given by:

${eV}_{0} = \frac{hc}{λ} - ϕ \Rightarrow {eV}_{0} = \frac{12420 eV Å}{3300 Å} - 2 eV \Rightarrow V_{0} = 1.75 V$

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