Question

Electrons in hydrogen-like atoms $(Z=3)$ make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wave length is $3.95\text{V}$. The work function of the metal $=x\text{eV}$. Then the value of $x$ is _______.
$\text{Rydberg constant}=1.094\times {10}^7{\text{m}}^{-1}$

Answer 1

The stopping potential for shorter wavelength is $3.95\text{volt}$ i.e., maximum kinetic energy of photoelectrons corresponding to shorter wavelength will be $3.95\text{eV}$. Further energy of incident photons corresponding to shorter wavelength will be in transition from $n_i=4$ to $n_f=3$.

$E_{4-3}=E_4-E_3=\frac{-\left(13.6\right)(3{)}^2}{(4{)}^2}-\left[\frac{-\left(13.6\right)(3{)}^2}{(3{)}^2}\right]=5.95\text{eV}$

Now from the equation, $K_{\text{max}}=E-\varphi$

$\Rightarrow \varphi =E-K_{\text{max}}=E_{4-3}-K_{\text{max}}$

$\therefore \varphi =(5.95-3.95)\text{eV}=2\text{eV}$

Hence, the correct answer will be $2$.