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1.5 mW of 400 nm light is directed at a ...

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 the cell.

Text Solution

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Let the number of photons hitting the photocell per second be n , then
`nhv=1.5mW`
or `n=((1.5xx10^(-3))xx400xx10^(-9))/((6.63xx10^(-34))xx(3xx10^(8)))`
`[v=(c)/(lamda)=(3xx10^(8))/(400xx10^(-9))]`
`n=3xx10^(15)`
Number of photoelectrons produced per second is
`3xx10^(15)xx(0.1)/(100)=(3xx10^(12))`
So, current in the photocell is
`(3xx10^(12))(1.6xx10^(-19))=4.8xx10^(-7)A`
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