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 photo-electron is
`K_(max)=1/2mv_(max)^(2)=(hc)/lambda-phi_(0)= ((6.6xx10^(-34))xx(3xx10^(8)))/(180xx10^(-9))-2xx1.6xx10^(-19)=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 inducing B is given by
`Bev=(mv^(2))/r or r=(mv)/(eB)=(9.1xx10^(-31)xx1.3093xx10^(6))/(1.6xx10^(-19)xx5.0xx10^(-5))=0.419m`