Anode voltage is at +3V. Incident radiation has frequency 1.4×1015Hz and work function of the photo cathode is 2.8 eV. Find the minimum and maximum KE of photo electrons in eV
3, 6
0, 3
0, 6
2.8, 5.8
A
0, 6
B
3, 6
C
0, 3
D
2.8, 5.8
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Solution
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Energy of the incident radiation is: hf=6.625×10−34×1.4×10151.6×10−19=5.8eV
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