Light of wavelength 1800 `Å` ejects photoelectrons from a plate of a metal whose work functions is 2 eV. If a uniform magnetic field of `5xx10^(-5)` tesla is applied parallel to plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy.

Explanation

Energy of incident photons in eV is given as
E= `(12431)/(1800)eV`
As work function of metal is 2 eV, the maximum kinetic energy of ejected electrons is
`KE_("max") = E-phi`
`=6.9 - eV = 4.9 eV`
If `v_("max")` be the speed of fasted electrons then we have
`(1)/(2)mv_("max")^2 = 4.9xx1.6xx10^(-19)` joule
or
`v_("max") = sqrt((2xx4.9xx1.6xx10^(-19))/(9.1xx10^(-31))` = `1.31xx10^6` m/s
When an electron with this speed enters a uniform magnetic field normally it follows a circular path whose radius can be given by
r=`(mv)/(qB)` [As qv `B= (mv^2)/r]`
or r=`(9.1xx10^(-31)xx1.31xx10^(6))/(1.6xx10^(-19)xx5xx10^(-5))`
or r=0.149 m