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When the frequency of the incident light...

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]

Text Solution

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`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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