1.

1.5mW of400 nm light is directed at a photoelectric cell. If0.10% of the incident photonsproduce photelectmthe current in the cell is(A) 0.36 HA16.(B) 0.48 HA(C) 0.42 mA(D) 0.32 mA

Answer»

for wavelength 400 nm , energy of each photon be E=hc/lamda

n number of photons will have total energy n(E) ...

power of photon beam is given so

n(E) = 1.5mW .....

n = 1.5/E (number of photons striking per second)

0.1% photons extract electrons so

ne(number of ejected electrons) = 0.1% of n

now photo current = nq=ne (e is electronic charge)I =0.48microampere.



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