The ratio of acceleration of the electron in singly ionized helium atom and hydrogen atom according to bohr’s model (both in ground state) is  $x:1$. When a metallic surface is illuminated with monochromatic light of wavelength  $\lambda$, the stopping potential is V. When the same surface is illuminated with light of wavelength  $(x-5)\lambda$, the stopping potential is  $V/5$. Work function of the metallic surface is  $hc/y\lambda$. find  $x-y$.

In a Bohr’s atom.
 Orbital velocity,  $\text{v}\propto \frac{Z}{n}\mathrm{..........}\left(i\right)$
 Orbital radius,  $r\propto \frac{n^2}{Z}\mathrm{..........}\left(ii\right)$
 So, for circular motion of electron. Acceleration is given by
$a=\frac{v^2}{r}\propto \frac{{\left(\frac{Z}{n}\right)}^2}{\frac{n^2}{Z}}$   [using Eqs.(i) and (ii)]
  $\Rightarrow a\propto Z^3$
So, for hydrogen  $(Z_1=1)$  and helium  $(Z_2=2)$, ratio of corresponding electronic acceleration is
$\frac{a_2}{a_1}={\left(\frac{Z_2}{Z_1}\right)}^3={\left(\frac{2}{1}\right)}^3=8\Rightarrow x=8\mathrm{....}\left(iii\right)$
By Einstein’s equation, stopping potential (V) is given by
 $V=\frac{hc}{e\lambda }-\frac{\varphi }{e}\mathrm{..........}\left(iv\right)$
Here,  $\lambda =wavelengthofincidentradiation$,  $\varphi =workfunction$
h,c,e are constants.
So, for wavelength equal to
$(x-5)\lambda =(8-5)\lambda =3\lambda$ [Using Eq. (iii)] stopping potential according to Eq.  (iv) is
$\frac{V}{5}=\frac{hc}{e\left(3\lambda \right)}-\frac{\varphi }{e}$

$\Rightarrow V=\frac{5hc}{3e\lambda }-\frac{5\varphi }{e}\mathrm{......}\left(v\right)$

$$\begin{gathered} \Rightarrow \varphi =\frac{hc}{6\lambda }=\frac{hc}{y\lambda }\left(given\right) \\ \Rightarrow y=6\mathrm{.....}\left(iv\right) \end{gathered}$$