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When the frequency of the incident light on a photo sensitive metal is changed from 7.6 xx 10^(14) Hz to 6 xx 10^(14) Hz the value of stopping potential changes by 0.66 V. Calculate Planck's constant. [Note : Any form of Einstein's P.E. Equation can be considered] |
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Answer» Solution :`HV = hv_(0) + eV_(s)` `H xx 7.6 xx 10^(14) = hv_(0) + 1.6 xx 10^(-19)V_(1)` `h xx 6 xx 10^(14) = hv_(0) +1.6 xx 10^(-19)V_(2)` Subtracting the above two equations we GET `h(1.6 xx 10^(14)) = 1.6 xx 10^(-19)(V_(1)-V_(2))` Here `(V_(1)-V_(2)) = 0.66` given, simplifying we get `h = 6.6 xx 10^(-34)` Js |
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