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 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

(i)Radius of circular path followed by charged particle in moving perpendicular to the field is given by the following expression :
`r=(mv)/(eB) rArr v=(eBr)/m`
Hence, kinetic energy of electron can be written as follows:
K.E. =`1/2mv^2=1/2m((eBr)/m)^2=(e^2B^2r^2)/(2m)`
On substituting values we get the kinetic energy of particle as follows :
K.E.=`(e^2B^2r^2)/(2m)`
`rArr K.E. =((1.6xx10^(-19))^2(1/320)^2(10^(-3))^2)/(2xx(9.1xx10^(-31)))`
`rArr` K.E. = `(((1.6xx10^(-19))/320)^2(10^(-3))^2)/(2xx(9.1xx10^(-31)))`
`rArr` K.E.=`((10^(-19))^2(10^(-3))^2)/(4xx10^4xx2xx(9.1xx10^(-31)))`
`=0.0137xx10^(-19)` J
Or we can write the K.E. in eV also.
`rArr` K.E.=`(1.37xx10^(-19))/(1.6xx10^(-10))`eV =0.856 eV
(ii)Energy of electron in `n^(th)` level of hydrogen is given by the following expression :
`E_n=-13.6/n^2 eV`
`E_3=-13.6/9^2 eV`=-1.51 eV
`E_2=-13.6/2^2` eV =-3.4 eV
Energy of the photon emitted due to transition from n=3 to n=2 can be written as follows :
`hv=E_3-E_2=(-1.51)-(-3.4)`=1.89 eV
According to photoelectric equation we can write the following :
hv=W+K.E.
here, W is work function of the material .
`rArr` W=hv-K.E. =1.89-0.856=1.034 eV
(iii)We have energy of the emitted photon as 1.89 eV . Hence we can write the following :
`(hc)/lambda=1.89xx1.6xx10^(-19)`
`lambda=((6.63xx10^(-34))(3xx10^8))/(1.89xx1.6xx10^(-19))`
`=6.577xx10^(-7)m = 657.7xx10^(-9)` m
=657.7 nm