Ultraviolet light of wavelength `800 A` and `700 A` when allowed to fall on hydrogen atoms in their ground states is found to liberate electrons with kinetic energies `1.8eV` and `4.0eV`, respectively. Find the value of Planck's constant.

Explanation

The energy of incident photon `= (hc//lambda)`
If `W`, is the ionization energy and `E_(K)` the kinetic energy of the emitted electrons , then we have
`(hc)/(lambda_(1)) = W_(i) + E_(K)`
For incident of wavelength `lambda_(1) = 800 Å = 8 xx 10^(-8) m`,
`(hc)/(lambda_(1)) = W_(i) + E_(K_1)` ...(i)
and for incident of wavelength `lambda_(2) = 700 Å = 7 xx 10^(-8) m`
`(hc)/(lambda_(2)) = W_(i) + E_(K_2)` ...(ii)
Subtracting Eq. (i) from Eq (ii) we get `((1)/(lambda_(2))/(1)/(lambda_(1))) = E_(K_2) - E_(K_1)`
or `h = (( E_(K_2) - E_(K_1)) lambda_(1) lambda_(2))/(c (lambda_(1) - lambda_(2)))`
Hence, `E_(K_1) = 1.8 e V xx 1.6 xx 10^(-19) J`
and `E_(K_2) = 4.0 e V xx 4.0 xx 10^(-19) J`
Subtracting the given values, we get
`h = ((4.8 - 1.8) xx 1.6 xx 10^(-19) xx 8 xx 10^(-8) xx 7 xx 10^(-8))/(3 xx 110^(8) (8 xx 10^(-8) - 7 xx 10^(-8)))`
`- 6.57 xx 10^(-34) J - s`