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Q.

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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a

0.75 V

b

1.75 V

c

2.5 V

d

3.75 V

answer is B.

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Detailed Solution

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

eV0 = hcλ - ϕ  ⇒eV0 = 12420 eV Å3300 Å - 2 eV ⇒V0 = 1.75 V

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