Question

Radiation corresponding to the transition $n=4$ to $n=2$ in hydrogen atoms falls on a certain metal (work function $=2.5eV$). The maximum kinetic energy of the photo-electrons will be:

• A. $0.55eV$
• B. $2.55eV$
• C. $4.45eV$
• D. none of these

Answer 1

The correct option is A $0.55eV$

Radiation corresponding to the transition

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{\lambda }=R{Z}^2\frac{1}{2^2}-\frac{1}{4^2}$

$\frac{1}{\lambda }=R{Z}^2\frac{3}{16}$

$R=109678c{m}^{-1}$ and $Z=$ Atomic no. of atom

$\frac{1}{\lambda }=109678\times 1^2\times \frac{3}{16}=20564c{m}^{-1}$

$\lambda =\frac{1}{20564}cm$

$K{E}_{m}ax=\frac{hc}{\lambda }-\Phi$

After putting the value of \lambda, h and c in CGS

$K{E}_{m}ax=\frac{hc}{\lambda }-2.5ev=0.55ev$