The rediation 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

The wavelength of emitted radiation is given as (i) `(1)/(lambda)=z^(2)R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`,For the lithium atom, Z=3,`therefore (1)/(lambda)=3^(2)R ((1)/(3^(2))-(1)/(4^(2)))=9R xx(7)/(144)` (ii) Radius of circular path of the electron in the magnetic field `r=(mv)/(qB)`,`v=(qBr)/(m)=1.18xx10^(6)m//s`,Maximum KE of the electron =`(1)/(2)mv^(2)=3.96eV`,(iii) `KE_(max)=(hc)/(lambda)-phi`,`phi =(hc)/(lambda)-KE_(max)=(5.96-3.96)=2 eV`