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. `6.0xx10^(-34)`
B. `6.4xx10^(-34)`
C. `6.6xx10^(-34)`
D. `6.8xx10^(-34)`

Answer

B

Explanation

`K.E._(max)=(hc)/lamda-phi=eV_0`
We can write the above equation for two different cases and then subtract from each other . We get the following equation :
`(hc)/lamda_1 -(hc)/lamda_2=e(V_1-V_2)`
`hc(1/(0.3)-1/(0.4))1/(10^(-6))=1.6xx10^(-19)`
`hc((0.1)/(0.12))=1.6xx10^(-25)`
`h=(1.6xx10^(-25)xx1.2)/(3xx10^8)`
`=0.64 xx10^(-33) = 6.4xx10^(-34) Js.`
We can calculate and verify the same using different values . We shall get the same result.