The rediation emitted when an electron jumps from `n = 3 to n = 2` orbit in a hydrogen atom falls on a metal to produce photoelectron. The electron from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of `(1//320) T` in a radius of `10^(-3) m`. Find (a) the kinetic energy of the electrons, (b) Work function of the metal , and (c) wavelength of radiation.

Explanation

a. When charges particle moves perpendicular to magnetic firld, then magnetic field provides necessary centripetal force for circular path of radius r given by ltBrgt `(mv^(2))/(r)=qvBimpliesm upsilon=qBr`
As momentum, `p=mv=sqrt(2mE_(k))`
where `E_(k)` is kinetic energy
`sqrt(2mE_(k))=qBr`
`E_(k)=((qBr)^(2))/(2m)`
`=({1.6xx10^(-19)xx((1)/(320))xx10^(-3)}^(2))/(2xx9.1xx10^(-31))J`
`=1.374xx10^(-19)J`
`=(1.374xx10^(-19))/(1.6xx10^(-19))eV=0.86eV`
b. Energy of photon relesed due to trasiton from `n=3` to `n=2` in hydrogen atom,
`epsi=triangleE`
`=Rhc((1)/(n_1^2)-(1)/(n_2^2))`
`=(13.6eV)((1)/(2^(2))-(1)/(3^(2)))=1.89eV`
Work function of metal,
`W=epsiE_k`
`=1.89-0.86=1.03eV`
c. wavelength of emitted radiation (photon) is given by `triangleE=(hc)/(lamda)`
`implies=lamda=(hc)/(triangleE)=(6.62xx10^(34)xx3xx10^(8))/(1.89xx1.6xx10^(-19))`
`=6.567xx10^(-7)m=6567A`