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One milliwatt of light of wavelength 456...

One milliwatt of light of wavelength 4560 A is incident on a cesium surface. Calculate the photoelectric current liberated assuming a quantum efficiency of 0.5 %. Given Planck's constant `h=6.62xx10^(-34)J-s` and velocity of light `c=3xx10^(8)ms^(-1)`.

Text Solution

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The energy of each photon of incident light is
`E=hv=(hc)/(lamda)=((6.62xx10^(-34))(3xx10^(3)))/((4560xx10^(-10)))`
`=4.34xx10^(-19)J`
Number of photons in one milliwatt source
`=(10^(-3))/(4.35xx10^(-19))=2.29xx10^(15)s^(-1)`
As quantum efficiency `=0.5%`. Hence, number of electrons liberated per second `=2.29xx10^(15)xx(0.5)/(100)`
`=1.14xx10^(13)` per second
photoelectric current `=(1.14xx10^(13))(1.6xx10^(-19))`
`=1.824xx10^(-16)A=1.824muA`
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