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Photoelectric threshold of silver is `lamda=3800A`. Ultraviolet light of `lamda=2600A` is incident of a silver surface.(Mass of the electron `9.11xx10^(-31)kg`)` 1. Calculate the value of work function is eV.A. 1.77B. 3.27C. 5.69D. 2.32 |
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Answer» Correct Answer - B `lamda_0=3800A` `W=hf_0=h(c)/(lamda_0)=(6.633xx10^(-34)xx3xx10^(8))/(3800xx10^(-10))` `=5.23xx10^(-19)J=3.27eV` Incident wavelength `lamda=2600A` `f=` incident frequency `=(3xx10^(8))/(2600xx10^(-10))Hz` Then, `KE_(max)=hf-W-0` `hf=(6.63xx10^(-34)xx3xx10^(8))/(2600xx10^(-10))` `=7.65xx10^(-19)J=4.78eV` `KE_(max)=hf-W-0=4.78eV-3.27eV=1.51eV` `KE_(max)=(1)/(2)mv_(max)^(2)` or `v_(max)=sqrt((2KE_(max))/(m))` |
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