Radiation of wavelength 180 nm eject photoelectrons from a plate whose work function is 2.0 eV. If a uniform magnetic field of flux density `5.0xx10^(-5)T` is applied parallel to plate, what should be the radius of the path followed by electrons ejected normally from the plate with maximum energy:

Options

A. 0.074 m
B. 0.592m
C. 0.419m
D. 0.149

Answer

D

Explanation

`E-hv-w=(hc)/(lambda)-w`
Here `lambda= nm=180xx10^(-19)m`
`w=2*0eV=2*0xx1*6xx10^(-9)J`
`:.E=7*8xx10^(-19)J`
Also `(1)/(2) mv^(2)=E:. V= sqrt((2E)/(m))`
So `v=sqrt((2xx7*8xx10^(-19))/(9*1xx10^(-31)))=1*31xx10^(6)m//s`
Radius r in a magnetic field of induction B is given by
`Bev=(mv^(2))/(r) or r=(mv)/(eB)`
Using `B=5xx10^(-5)T`
Then `r=0*149m`