Light of wavelength 180 nm ejects photoelectrons from a plate of metal whose work - function is 2 eV. If a uniform magnetic field of `5xx10^(-5)` T be applied parallel to the plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy.

Answer

`0.149 cm`

Explanation

Here, `lambda=180xx10^(-19)m=1.80xx10^(-7)m`,
`B=5.0xx10^(-5)T`,
`phi_(0)=2.0eV=2xx1.6xx10^(-19)J`
`=3.2xx10^(-19)J`
As, `1/2mv^(2)=(hc)/lambda-phi_(0)`
`=(6.63xx10^(-34)xx3xx10^(8))/(1.8xx10^(-7))-3.2xx10^(-19)`
`=11xx10^(-19)-3.2xx10^(-19)=7.8xx10^(-19)J`
`v=sqrt((2xx7.8xx10^(-19))/m)=sqrt((2xx7.8xx10^(-19))/(9.1xx10^(-31)))`
`=1.039xx10^(6)ms^(-1)`
The electron moving in a perpendicular magnetic field describes a circular path of radius r, then
`evB=(mv^(2))/r`
or `r=(mv)/(eB)=((9.1xx10^(-19))xx(1.309xx^(6)))/((1.6xx10^(-19))xx(5.0xx10^(-5)))`
`=1.49xx10^(-3)m=1.49mm`