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Question

The work function of a metal is 3.0eV. It is illuminated by a light of wave length 3×107m. then
i) Threshold frequency
ii) maximum energy of photoelectrons,
iii) the stopping potential.

(h=6.63×1034mJs and c=3×108m/s)

A
0.72×1015Hz,1.81×1019J,1.13V
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B
0.42×1015Hz,1.96×1019J,2.13V
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C
0.89×1015Hz,3.92×1019J,6.14V
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D
1.67×1015Hz,2.3×1019J,3.1V
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Solution

The correct option is A 0.72×1015Hz,1.81×1019J,1.13V
Threshold frequency =ϕh

=3×1.6×10196.63×1034

=0.72×1015HZ.

maximum energy =hcλϕ

=12403003 (hc=1240evnm)

=1.13ev.
=1.13×1.6×1019
=1.81×1019J.
So, stopping potential
=maximum energy/e
=1.13ev/e
=1.13v
So, the answer is (A).

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