In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength `(lambda)`. ) of incident light and the corresponding stopping potential `(V_0 )` are given below :

Given that `c=3xx10^8ms^(-1) and e=1.6xx10^(-19)C` Planck's constant (in units of J s) found from such an experiment is

Options

A. `6.0 xx 10^(-34)`
B. `6.4 xx 10^(-34)`
C. `6.6 xx 10^(-34)`
D. `6.8 xx 10^(-34)`

Answer

B

Explanation

`KE_("max") = (hC)/(lambda) - phi`
`eV_(s) = (hC)/(lamda) - phi`
`1.6 xx 10^(-19) xx 2 = (h xx 3 xx 10^(8))/(3000 xx 10^(-10)) - phi`.....(i)
`1.6 xx 10^(-19) xx 1 = (h xx 3 xx 10^(8))/(4000 xx 10^(-10)) - phi`......(ii)
From (ii) `phi = (h xx 3 xx 10^(8))/(4000 xx 10^(-10)) - 1.6 xx 10^(-19)`
`1.6 xx 10^(-19) xx 2 = (h xx 3 xx 10^(8))/(3000 xx 10^(-10)) - (h xx 3 xx 10^(8))/(4000 xx 10^(-10)) + 1.6 xx 10^(-19)`
`1.6 xx 10^(-19) = (h xx 3 xx 10^(8))/(10^(-7)) ((1)/(3) - (1)/(4)) = (h xx 3 xx 10^(8))/(10^(-7)) [(4 - 3)/(12)]`
`1.6 xx 10^(-19) = (h xx 3 xx 10^(8))/(10^(-7)) xx (1)/(12)`
`1.6 xx 4 xx (10^(-19) xx 10^(-7))/(10^(8)) = h`
`6.4 xx 10^(-34) Js = h`