Question

(a) Calculate the energy of an excited state for hydrogen.

(b) Define threshold frequency of work function.

(c) Work function for $C_5$ atom is 1.9eV. Calculate kinetic energy for photoemitted electron if $C_5$ metal is irradiated with radiation of 500nm.

Answer 1

Centripetal force is provided by electrostatic art

$\frac{m{v}^2}{r}=+\frac{Kexe}{r^2}$

$\Rightarrow m{v}^2=+\frac{K{e}^2}{r}\Rightarrow v^2=\frac{K{e}^2}{rm}$ ...(i)

$\therefore KE=\frac{1}{2}m{v}^2=+\frac{K{e}^2}{2r}$

Also,

$P>E=\frac{K{q}_1{q}_2}{r}$

$=\frac{K\times e\times (-e)}{r}$

$=-\frac{K{e}^2}{r}$

$TE=-\frac{K{e}^2}{2r}$ $(PE+KE)$ ...(ii)

Also,

$mvr=\frac{nh}{2\pi }$ (Bohr's part)

$\Rightarrow m^2{v}^2{r}^2=\frac{n^2{h}^2}{4{\pi }^2}$

$\Rightarrow m^2{r}^2\times \frac{K{e}^2}{rm}=\frac{n^2{h}^2}{4{\pi }^2}$ (from 1)

$\Rightarrow r=\frac{n^2{h}^2}{4{\pi }^{2}ml{e}^2}$

$TE=-\frac{K{e}^2}{n^2{h}^2}\times 4{\pi }^{2}mK{e}^2=-\frac{K^2{e}^{4}4{\pi }^{2}m}{n^2{h}^2}$

(b) Threshold frequency is the min. frequency required by a proton to eject a electron out of the element

(c) $E_0=1.9eV$ $E=\frac{hc}{\lambda }=\frac{6626\times {10}^{-34}\times 3\times {10}^8}{500\times {10}^{-5}\times 16\times {10}^{-19}}eV$

$=2.48eV$

$\therefore KE=2.48-1.9$

$=0.58eV$