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The work function of Cs is 2.14 eV. Find (a) threshold frequency for Cs. (b) Wavelength of incident light if the photo current is brought to zero by stopping potential of 0.6V. |
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Answer» Solution :(a) As work function `phi_(0)=2.14eV=2.14xx1.6xx10^(-19)J` `therefore"Threshold FREQUENCY "v_(0)=(phi_(0))/(h)=(2.12xx1.6xx10^(-19))/(6.63xx10^(-34))=5.17xx10^(14)Hz` (b) If stopping potential `V_(0)=0.6V`, then energy of incident photon `E=(hc)/(lambda)=phi_(0)+eV_(0)` `=2.14eV+0.6eV=2.74eV_(0)=2.74xx1.6xx10^(-19)J` `rArr""lambda=(hc)/(E)=((6.63xx10^(-34))xx(3xx(10^(8))))/((2.74xx1.6xx10^(-19)))=4.53xx10^(-7)m=453nm` |
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