the radiation emitted when an electron jumps from `n=4` to `n=3` in a lithium atom (`z=3` ) fall on a metal surface to produce photoelectron , when the photoelectron with maximum Kinetic energy are made to move perpendicular to a magnetic field of `2xx10^(-4)T` , it traces a circular path of radius `sqrt(9.1/(1.6))cm`, [RhC=13.6eV] (mass of electron=`9.1xx10^(-31)kg)`

Options

A. the wavelength of radiation falling on metal is `208nm` (nearly)
B. the work function of metal is `3.95eV`
C. the kinetic energy of photoelectron is `6eV`
D. the energy of incident photonn is `5.95eV`

Answer

A, B, D

Explanation

Energy of emitted photons
` E=13.6z^(2)(1/n_(1)^(2)-1/n_(2)^(2))=13.6xx9(1/9-1/16)=5.95eV`
` thereforelambda=(hc)/E=(1240)/(5.95)=208.4nm`
` r=(mv)/(Bq)=sqrt(2mKE_(max))/(Bq)RightarrowKE_(max)=(B^(2)q^(2)r^(2))/(2m) =2eV`
`thereforeW=E-KE_(max)=3.95eV`