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 with maximum kinetic energy are made to move perpendicular to a magnetic field of `(1)/(320)` T in a radius of `10^(m)`. The work function of the metal is :

Options

A. 3.03 eV
B. 2.03 eV
C. 1.03 eV
D. 0.03 eV

Answer

A

Explanation

Wavelength of emitted radiation is given by
`(1)/(lambda)=R((1)/(2^(2)))-(1)/(3^(2))=(5R)/(36)`
Given `r=10^(-3) m and B=(1)/(320)T`
Now `r=(mv)/(eB)=(sqrt(2mk))/(eB)`
Here K is K.E. of emitted electron
`:.K=(r^(2)e^(2)B^(2))/(2m)`
`=(10^(-6)xx(1.6xx10^(-19))^(2)xx1)/(2xx(9.1xx10^(-31))^(2)(320)^(2))=(10^(-17))/(8xx9.1)`
`=(10^(-17))/(72.8xx1.6xx10^(-19))eV=0.86eV`
Now `(hc)/(lambda)=K+w`
`:.w=(hc)/(lambda)-K`
`=(5Rhc)/(36)-K`
`=(5xx13.6)/(36)eV=0.86eV`
` w=103.eV` is work function of metal