Question

The radiation emitted when an electron jumps from $n=4$ to $n=3$ in a $L{i}^{++}ion$ (atomic no. $Z=3$ ) falls on a metal surface to produce photoelectron. When the photoelectron with maximum kinetic energy is made to move perpendicular to a uniform magnetic field of $2\times {10}^{-4}T,$ it follows a circular path of radius $\sqrt{\frac{9.1}{1.6}}cm.$ The work function of metal (in eV ) is (given mass of electron $m=9.1\times {10}^{-31}kg\text{and charge on electron}=-1.6\times {10}^{-19}C$

Answer 1

$r=\frac{mv}{qB}=\frac{\sqrt{2m{K}_{max}}}{qB}$

$\Rightarrow K_{max}=\frac{q^2{B}^2{r}^2}{2m}$

$\text{Putting the values of q, b, r and m}{K}_{max}=2eV$

$\text{Energy of emitted photon}hv=13.6\times 3^2(\frac{1}{9}-\frac{1}{16})eV=5.95eVhv=\varphi +K_{max}\Rightarrow \varphi =3.95eV\approx 4eV$