Question

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 1

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`