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When a centimeter thick surface is illuminated with light of wavelength `lamda`, the stopping potential is V. When the same surface is illuminated by light of wavelength `2lamda`, the stopping potential is `(V)/(3)`. Threshold wavelength for the metallic surface isA. `(4lamda)/(3)`B. `4lamda`C. `6lamda`D. `(8lamda)/(3)` |
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Answer» Correct Answer - B `eV=hc((1)/(lamda)-(1)/(lamda_0))` `(eV)/(3)=hc((1)/(2lamda)-(1)/(lamda_0))` Dividing Eq. (i) and (ii), we get `lamda_0=4lamda` |
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