Question

The radiation, emitted when an electron jumps from $n=4$ to $n=3$ in a Lithium atom, falls on a metal surface to produce photoelectron. When the photoelectrons with maximum $K.E.$ are made to move perpendicular to a uniform magnetic field of $4\times {10}^{-4a}T$, it trace out a circular path of radius $1.68cm$. Find The radiation, emitted when an electron jumps from $n=4$ to $n=3$ in a Lithium atom, falls on a metal surface to produce photoelectron. When the photoelectrons with maximum $K.E.$ are made to move perpendicular to a uniform magnetic field of $4\times {10}^{-4a}T$, it trace out a circular path of radius $1.68cm$. Find the
(A) a wavelength of radiation falling on the metal surface.
(B) the kinetic energy of electrons.
(C) a work function of the metal.

Answer 1

The wavelength of emitted radiation is given as
$\frac{1}{\lambda }=z^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$,
For the Lithium atom, $Z=3$
$\therefore \frac{1}{\lambda }=3^{2}R(\frac{1}{3^2}-\frac{1}{4^2})=9R\times \frac{7}{144}$
$\lambda =2.08\times {10}^{-7}$
Radius of circular path of the electron in the magnetic field
$r=\frac{mv}{qB}$
$v=\frac{qBr}{m}=1.18\times {10}^{6}m/s$
Maximum $KE$ of the electron $=\frac{1}{2}m{v}^2=3.96eV$
$K{E}_{max}=\frac{hc}{\lambda }-\varphi$
$\varphi =\frac{hc}{\lambda }-K{E}_{max}=(5.96-3.96)=2eV$