The radiation emitted when an electron jumps from `n = 3 to n = 2` orbit in a hydrogen atom falls on a metal to produce photoelectron. The electron from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of `(1//320) T` in a radius of `10^(-3) m`. Find (a) the kinetic energy of the electrons, (b) Work function of the metal , and (c) wavelength of radiation.

Explanation

When charged particle moves perpendicular to a magnetic field , then magneticfield provides necessary centripetel force for circle path of radius`r` given by `(m v^(2))/( r) = q v B`
`implies m v = q B r`
As momentum, `p = mv = sqrt(2 m E_(K))`, where `E_(K)` is kinetic energy , therefore
`sqrt(2 m E_(K)) = qBr`
`:. E_(k) = ((qBr)^(2))/(2 m)`
`{1.6 xx 10^(-19) xx ((1)/(320)) xx 10^(-3)}^(2)/(2 xx 9.1 xx 10^(-31) J`
`= 1.374 xx 10^(-19)J = (1.374 xx 10^(-19))/(1.6 xx 10^(-19)) e V`
`= 0.86 e V`
Energy of photon released due to transition from `n = 3 to n = 2` in hydrogen atom.
`epsilon = Delta E`
`= Rhc ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = (13.6 e V) ((1)/(2^(2)) - (1)/(3^(2)))`
`= 1.89 e V`
Work function of metal `W = epsilon - E_(k) = 1.89 - 0.86 = 1.03 e V`
c. Wavelength of emitted radition (photon) is given by
`Delta E = (hc)/(lambda)`
`implies lambda = (hc)/(Delta E) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(1.89 xx 1.6 xx 10^(-19))`
`= 6.567 xx 10^(-7) m = 6567Å`