Question

A gas containing hydrogen like ions with atomic no. Z, emits photons in transition $n+2\to n$, where $n=Z$. These photons fall on a metallic plate and eject electrons having minimum de-Broglie wavelength $\lambda$ of $5A^{\circ }$. Find the value of $‘Z^{\prime }$ if the work function of metal is $4.2\text{eV}$.

Answer 1

The energy of transition can be written as-
$E=-\left(13.6\text{eV}\right)(\frac{1}{(Z+2{)}^2}-\frac{1}{Z^2})$
$E=-13.6\times \left(\frac{Z^2-(Z+2{)}^2}{(Z+2{)}^2}\right)$
$E=\frac{4(Z+1)\times 13.6}{(Z+2{)}^2}$
Now energy of electron is
$K=\frac{h^2}{2{\lambda }^{2}m};(\text{we have}\lambda =\frac{h}{\sqrt{2mK}})$
$K=6\text{eV}$

Using $E=K+\varphi$
$E=10.2\text{eV}$

So, $\frac{4(Z+1)\times 13.6}{(Z+2{)}^2}=10.2\text{eV}$

$\frac{(Z+1)}{(Z+2{)}^2}=\frac{3}{16}\Rightarrow (Z-2)(3Z+2)=0$

Neglecting the negative /fractional value $Z=2$