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The electric field associated with a monochromataic beam of light becomes zero, `2.4xx10^(15)` times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0eV, `h=6.63xx10^(-34)Js`. |
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Answer» Correct Answer - `2.97 eV` In one complete vibration twice the electric field becomes zero, so frequency of the incident light, `v=1/2xx2.4xx10^(15)=1.2xx10^(15)cps`. Max K.E. `=hv-phi_(0)` `=(6.63xx10^(-34)xx1.2xx10^(15))/(1.6xx10^(-19))-2.0` `=2.97 eV` |
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