Question

The frequency of emission line for any transition in positronium atom (consisting of a positron and an electron) is ${}^{\prime }x{}^{\prime }$ times the frequency for the corresponding line in the case of $H$ atom, where $x$ is

• A. $\sqrt{2}$
• B. $1/2\sqrt{2}$
• C. $1/\sqrt{2}$
• D. $1/2$

Answer 1

Answer: (D)

$1/2$

Energy levels in positronium is given by: $E_n=-\frac{6.8}{n^2}eV$
whereas, for hydrogen atom is, $E_n=-\frac{13.6}{n^2}eV$
Frequency of lines emitted is obtained by difference in energy levels

I) For Positronium atom is:
$E_2-E_1=h{\nu }_p=-6.8\times -3/4$

II) For Hydrogen atom is:
$E_2-E_1=h{\nu }_h=-13.6\times -3/4$

So we have, $\frac{{\nu }_p}{{\nu }_h}=\frac{6.8}{13.6}$

Hence, $x=\frac{6.8}{13.6}=\frac{1}{2}$