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 wavelength is 3.95 volts. Calculate the work function of the metal, and the stopping potential(in V) for the photoelectrons ejected by the longer wavelength.
$(Rydberg constant = 1.094 \times 10^{7} m^{- 1})$

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Solution

The energy of the $(n - 1) t h$ excited state is given by

$E = - 13.6 \frac{n^{2}}{Z^{2}}$

Therefore, the released radiations have energy

$Δ E_{1} = - 13.6 Z^{2} (\frac{1}{5^{2}} - \frac{1}{4^{2}}) = 2.75 e V$

$Δ E_{2} = - 13.6 Z^{2} (\frac{1}{4^{2}} - \frac{1}{3^{2}}) = 5.95 e V$

Recalling photoelectric effect,
$e V_{o} = h f - ϕ$
where $V_{o}$ is the stopping potential and $ϕ$ is the workfunction.
It is given that $V_{o 2} = 3.95 V$
$e V_{o 2} = Δ E_{2} - ϕ$

⟹ $ϕ = 5.95 - 3.95 = 2 e V$
∴ $e V_{o 1} = Δ E_{1} - ϕ$
⟹ $e V_{o 1} = 2.75 - 2 = 0.75 e V ⟹ V_{o 1} = 0.75 V$

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