In a historical experiment to dtermine 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 wavelength `(lamda)` of incident light and the corresponding stopping potential `(V_(0))` are given below

Given that `c=3xx10^(8)ms^(-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

For photoelectronic emission,
`(hc)/(lamda)=E_(0)+eV_(0)` . . . .(i)
Here, kinetic energy of photoelectron `((1)/(2)mv^(2))=eV_(0)`
`(hxx3xx10^(8))/(0.4xx10^(-6))=E_(0)+1.6xx10^(-19)xx2` . . . .(ii)
`(hxx3xx10^(8))/(0.4xx10^(-6))=E_(0)+1.6xx10^(-19)xx1` . . . .(iii)
Substracting eq. (iii) from eq. (ii) we get
`h[10^(15)-0.75xx10^(15)]=1.6xx10^(-19)` ltBrgt `hxx0.25xx10^(15)=1.6xx10^(-19)`
`h=6.4xx10^(-34)`Js