Physics - Modern Physics Question with Solution | TestHub

The energy of a photon is $E=hc/\lambda$. For He+ (Z=2), the energy levels are $E_n=-13.6\times Z^2/n^2$ eV. The total energy emitted is the sum of the energies of the two photons. Calculate the initial energy level $n$ using the total energy difference.

Given wavelengths: ${\lambda }_1=108.5$ nm and ${\lambda }_2=30.4$ nm.

Energy of emitted photons:

$E_1=\frac{1240eVnm}{108.5nm}=11.43eV$

$E_2=\frac{1240eVnm}{30.4nm}=40.79eV$

Total energy emitted: $E_{total}=E_1+E_2=11.43+40.79=52.22eV$.

This energy corresponds to the transition from an excited state $n$ to the ground state ($n_f=1$).

$E_{total}=E_n-E_1=-13.6\times Z^2(\frac{1}{n^2}-\frac{1}{1^2})$

$52.22=-13.6\times 2^2(\frac{1}{n^2}-1)$

$52.22=-54.4(\frac{1}{n^2}-1)$

$-0.96=\frac{1}{n^2}-1$

$0.04=\frac{1}{n^2}$

$n^2=25$

$n=5$