Question

In an experiment on the photo-electric effect, singly ionized helium is excited electronically to different energy levels. The light emitted by ionized helium is incident on a photo-electric plate in a photocell. When helium is excited to fourth energy level, then the observed stopping potential of the photocell is found to be five times the stopping potential measured when the photoelectrons are produced by using light emitted by hydrogen atom, excited to the third energy level. Determine the work function of the material of the photo-electric plate.

Answer 1

Energy of the photon with the highest frequency in the emission spectrum is
For helium (excited to the $4^{th}$ level):$f_{‘1}={\left(2\right)}^{2}Rc[\frac{1}{1^2}-\frac{1}{4^2}]$ $\left(1\right)$
For hydrogen (excited to the $3^{rd}$ level):$f_{‘2}=Rc[\frac{1}{1^2}-\frac{1}{3^2}]$ $\left(2\right)$
Using Einsteins equation of the photo-electric effect, we obtain,
$h{f}_1-W=e{V}_1$ $\left(3\right)$
$h{f}_2-W=e{V}_2$ $\left(4\right)$
Where $V_1=5V_2$ (given) $\left(5\right)$
Using $\left(3\right),\left(4\right)$ and $\left(5\right)$ we obtained,
$\frac{h{f}_1-W}{e}=5\left(\frac{h{f}_2-W}{e}\right)\Rightarrow 4W=h(5f_2-f_1)$
$\Rightarrow W=\frac{h}{4}[5f_2-f_1]$
Putting the values of $f_{1}and{f}_2$ from $\left(1\right)$ and $\left(2\right)$ in $\left(6\right)$ we obtain,
$W=\frac{h}{4}[(1-\frac{1}{9})5Rc-4(1-\frac{1}{16})Rc]=\frac{h}{4}Rc[\frac{8\times 5}{9}-\frac{15}{4}]=0.694Rhc$
Putting $Rhc=13.6eV$ we obtain,
$W=2.36eV$