Photoelectric effect and study of photoelectric effect

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Question

1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1 % of the incident photons produce photoelectrons, find the current (in $μ$ A) in the cell.

Moderate

Solution

Number of photons incident per second
$\begin{array}{l} = \frac{Power}{Energy of one photon} \\ = \frac{P}{(hc / λ)} = \frac{Pλ}{hc} \end{array}$
Number of electrons emitted per second = 0.1% of
$\frac{Pλ}{hc} = \frac{Pλ}{1000 hc}$
$∴$ Current = Charge (on photoelectrons per second)
$\begin{array}{l} = \frac{Pλe}{1000 hc} \\ = \frac{(1 .5 \times 10^{- 3}) (400 \times 10^{- 9}) (1 .6 \times 10^{- 19})}{(1000) (6 .63 \times 10^{- 34}) (3 \times 10^{8})} \\ = 0 .48 \times 10^{- 6} A = 0 .48 μA \end{array}$

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Photoelectric effect and study of photoelectric effect

Remember concepts with our Masterclasses.

1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1 % of the incident photons produce photoelectrons, find the current (in μA) in the cell.

Number of photons incident per second = Power Energy of one photon =P(hc/λ)=PλhcNumber of electrons emitted per second = 0.1% of Pλhc=Pλ1000hc∴Current = Charge (on photoelectrons per second) =Pλe1000hc=1.5×10−3400×10−91.6×10−19(1000)6.63×10−343×108=0.48×10−6A=0.48μA

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1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1 % of the incident photons produce photoelectrons, find the current (in $μ$ A) in the cell.