Energy levels of hydrogen

For the hydrogen atom, $E_n = -\frac{13.6}{n^2}\,\text{eV}$

For the transition from $n=4$ to $n=1$: $E_4 = -\frac{13.6}{16} = -0.85\,\text{eV}$ $E_1 = -13.6\,\text{eV}$

Energy of emitted photon

The emitted photon has energy equal to the difference: $\Delta E = E_4 - E_1 = (-0.85)-(-13.6)=12.75\,\text{eV}$

Use photon relation

$E = \frac{hc}{\lambda}$ So, $\lambda = \frac{hc}{E}$

Given: $h = 4\times 10^{-15}\,\text{eV s}, \quad c = 3\times 10^8\,\text{m/s}$

Therefore, $hc = 4\times 10^{-15}\times 3\times 10^8 = 12\times 10^{-7} = 1.2\times 10^{-6}\,\text{eV m}$

Hence, $\lambda = \frac{1.2\times 10^{-6}}{12.75}\,\text{m}$ $\lambda \approx 9.41\times 10^{-8}\,\text{m}$

Convert to nm

Since $1\,\text{nm} = 10^{-9}\,\text{m}$, $\lambda = 9.41\times 10^{-8}\,\text{m} = 94.1\,\text{nm}$

Check options

• A: $99.3\,\text{nm}$
• B: $94.1\,\text{nm}$
• C: $974\,\text{nm}$
• D: $941\,\text{nm}$

The correct option is B.