Answer
The lower quantum number of the transition producing this emission is 1.
Work Step by Step
Light of wavelength 121.6 nm is emitted by a hydrogen
atom.
$E_{photon-emitted} = \frac{hc}{λ} = \frac{(4.136×10^{-12} eV s)(3\times10^{8} m/s)}{121.6\times10^{-9}m} = 10.203 eV$
We know that:
$E_{photon-emitted} = -13.6(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
or, $10.203eV = -13.6(\frac{1}{n_{2}^{2}} - \frac{1}{n_{2}^{2}})$
or, $-0.75 \approx (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
or, $(\frac{1}{2^{2}}-\frac{1}{1^{2}}) = (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Comparing this, we obtain:
$n_{1} = 2$ and $n_{2} = 1$
Therefore, the lower quantum number of the transition producing this emission is 1.
