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.

Explanation

Max. K.E. of the ejected photoelectons is
`K_(max)=1/2mv_(max)^(2)=(hc)/lambda-phi_(0)`
`=((6.62xx10^(-34))xx(3xx10^(8)))/(180xx10^(-9))-2xx1.6xx10^(-10)`
`=7.8xx10^(-19)J`
`v_(max)=sqrt((2K_(max))/m)=sqrt((2xx7.8xx10^(-19))/(9.1xx10^(-31)))`
`=1.3093xx10^(6)m//s`
Radius of circular path in the magnetic field of induction B is given by , `Bev=mv^(2)//r`
or `r=(mv)/(eB)=((9.1xx10^(-31))xx(1.3093xx10^(6)))/((1.6xx10^(-19))xx(5.0xx10^(-5)))`
`=0.149m=14.9cm`