Ultraviolet light of wavelength `lambda_(1)` and `lambda_(2)` (with `lambda_(2) gt lambda_(1)`) when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies `E_(1)` and `E_(2)` respectively. The value of the planck's constant can be found from the relation

Options

A. `h=1/c(lambda_(2)-lambda_(1))(E_(1)-E_(2))`
B. `h=1/c(lambda_(2)+lambda_(1))(E_(1)+E_(2))`
C. `h=((E_(1)-E_(2))lambda_(1)lambda_(2))/(c(lambda_(2)-lambda_(1)))`
D. `h=((E_(1)+E_(2))lambda_(1)lambda_(2))/(c(lambda_(2)+lambda_(1)))`

Answer

C

Explanation

`:'E=hv=(hc)/(lambda)`
The energies of radiations of wavelengths `lambda_(1)` and `lambda_(2)` are
given by `E_(1)=hv_(1)=(hc)/(lambda_(1))` and `E_(2)=(hc)/(lambda_(2))`.
`:'lambda_(2) gt lambda_(1), E_(1) gt E_(2)`
`:.E_(1)-E_(2) = hc((1)/(lambda_(1))-(1)/(lambda_(2)))=(hc(lambda_(2)-lambda_(1)))/(lambda_(1)lambda_(2))`
`:.h=((E_(1)-E_(2))lambda_(1)lambda_(2))/(c(lambda_(2)-lambda_(1)))`