The radiation emitted when an electron jumps from `n=3` to `n=2` orbit of hydrogen atom falls on a metal to produce photoelectrons. The electrons emitted from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of `(1)/(320)` T is a radius of `10^-3m`. Find:
A. the kinetic energy of the electrons,
B. work function of metal, and
C. wavelength of radiation.
(Planck's constant `h=6.62xx10^-32J-s`)

Explanation

a. Radius of the circle traced by a charged particle in a magnetic field is given by
`R = (m v)/(B q) = (p)/(B q) = (sqrt2 mK)/(B q)`
Hence , `K = ((BqR)^(2))/(2 m) = [(10^(-3) xx 1.6 xx 10^(-19) xx(1//320)]^(2))/(2 xx 9.1 xx 10^(-31))`
`= (10^(-17))/(72.8) j = (10^(-17))/(72.8) xx (1)/(1.6 xx 10^(-19)) e V`
`= 0.86 e V`
b. Energy of the photon emitted when the electron makes a transition from `n = 3 to n = 2` energy level is
`h v = E_(3) - E_(2) = 13.6 [(1)/(2^(2)) - (1)/(3^(2))] e V = 1.9 e V`
From Einstein's photoelectric effect equation,
`h v - phi + KE_(max)`
So `phi = (1.9 - 0.86) e V = 1.04 e V`
c. wevelength of radition is
`lambda = ((6.63 xx 10^(-34)) (3 xx 10^(8)))/(1.9 xx 1.6 xx 10^(-19)) m = 6542 Å`