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

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`