Question

Consider a hypothetical hydrogen like atom. The wavelength in $A^o$ for the spectral lines for transition from $n=p$ to $n=1$ are given by:
$\lambda =\frac{1500p^2}{p^2-1},$ where $p>1$
Find the ionization potential of this element?

• A. $16.56$ eV
• B. $8.28$ eV
• C. $4.14$ eV
• D. $22.4$ eV

Answer 1

The correct option is D $22.4$ eV

The expression for the wavelength of the emitted radiation when the transition is from p to 1 is
$\frac{1}{\lambda }=109678Z^2(\frac{1}{1^2}-\frac{1}{p^2})=109678Z^2\left(\frac{p^2-1}{p^2}\right)$
$\lambda =\frac{1}{109678Z^2}\frac{p^2}{p^2-1}$......(1)
However the given expression is
$\lambda =1500\times {10}^{-8}\times \frac{p^2}{p^2-1}$......(2)
(Note : in the above expression, the Angstrom units are converted to cm units)
From equations (1) and (2)
$1500\times {10}^{-8}=\frac{1}{109678Z^2}$
$Z^2=1.65$
The ionization energy is $13.6Z^2=13.6\times 1.65=22.4eV$