The work function of a metal is 3.0eV. It is illuminated by a light of wave length 3×10−7m. then i) Threshold frequency ii) maximum energy of photoelectrons, iii) the stopping potential.
(h=6.63×10−34mJs and c=3×108m/s)
A
0.72×1015Hz,1.81×10−19J,1.13V
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B
0.42×1015Hz,1.96×10−19J,2.13V
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C
0.89×1015Hz,3.92×10−19J,6.14V
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D
1.67×1015Hz,2.3×10−19J,3.1V
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Solution
The correct option is A0.72×1015Hz,1.81×10−19J,1.13V Threshold frequency =ϕh
=3×1.6×10−196.63×10−34
=0.72×1015HZ.
maximum energy =hcλ−ϕ
=1240300−3(hc=1240ev−nm)
=1.13ev. =1.13×1.6×10−19 =1.81×10−19J. So, stopping potential =maximumenergy/e =1.13ev/e =1.13v So, the answer is (A).