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The K.E. of photo electron ejected from a metal surface by light of wave length 2000 Å range from zero to 3.2 × 10^–19 J. The stopping potential will be equal to

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2 V

3.2 V

1.6 V

6.4 V

answer is A.

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Detailed Solution

If kinetic energy is

$3.2 \times 10^{- 19} J then eV it is \frac{3.2 \times 10^{- 19}}{1.6 \times 10^{- 19}} eV = 2 eV$

Stopping potential =2 V

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