The allowed electron energies predicted by the Bohr model of the hydrogen atom are correct.(a) Determine the three lowest. (b) The electron can "jump" from a higher to lower energy. with a photon carrying away the energy difference. From the three energies found in part (a), determine three possible wavelengths of light emitted by a hydrogen atom.

The electron energy in a particular energy level, as predicted by Bohr model, is given as-

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{\text{eV}}}{{\mathbf{n}}^{\mathbf{2}}}$

E is the energy of an electron and nis the principle quantum number.

The expression for the wavelength of the photon, emitted during the transition from higher to lower energy level, is given by,

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{1240}\mathbf{}\mathbf{}\mathbf{\text{nm}}\mathbf{·}\mathbf{\text{eV}}}{{\mathbf{E}}_{\mathbf{\text{n}}}\mathbf{-}{\mathbf{E}}_{\mathbf{\text{m}}}}$

Here, the photon is jumping from the m^thenergy level to n^thenergy level.

(a) The formula for the prediction of electron energy in a particular energy level, according to the Bohr model, is

$E_n=\frac{-13.6\text{eV}}{n^2}$

For energy level having $n=1$, the energy of the electron is –

$$\begin{gathered} E_1=\frac{-13.6\text{eV}}{1^2} \\ =-13.6\text{eV} \end{gathered}$$

For energy level having $n=2$,the energy of the electron is –

$$\begin{gathered} E_2=\frac{-13.6\text{eV}}{2^2} \\ =-3.4\text{eV} \end{gathered}$$

For energy level having $n=3$, the energy of the electron is –

$$\begin{gathered} E_3=\frac{-13.6\text{eV}}{3^2} \\ =-1.51\text{eV} \end{gathered}$$

Therefore, the three lowest electron energies are-$-13.6\text{eV},-3.4\text{eV and}-1.5\text{eV}$

(b)

When the electron jumps from a higher energy level to a lower energy level, it emits a photon. The energy of the photon emitted is equal to the energy difference between the two levels between whom transition is taking place. As we have three levels here, we will have only three possible transitions. The wavelength of the photons emitted in these transitions is given as-

$E_n-E_{n-1}=\frac{hc}{\lambda }··········································\left(1\right)$

Here, $h$is Planck’s constant and $c$is the speed of light in a vacuum.

So, equation (1) becomes,

localid="1660054696048" $\lambda =\frac{1240\text{eV}·\text{nm}}{E_n-E_{n-1}}$

For the transition, localid="1660054773541" $n=3\text{to}n=1$, the wavelength of photon is-

localid="1660055894008" $$\begin{gathered} {\lambda }_1=\frac{1240\text{eV.nm}}{-1.51\text{eV}-(-13.6)\text{eV}} \\ =\frac{1240\text{eV.nm}}{12.09\text{eV}} \\ =102.56\text{nm} \end{gathered}$$

For the transition, $n=2\text{to}n=1$,the wavelength of photon is-

localid="1660057033064" $$\begin{gathered} {\lambda }_2=\frac{1240\text{eV}.\text{nm}}{-1.51\text{eV}-(-3.4)\text{eV}} \\ =\frac{1240\text{eV}·\text{nm}}{1.89\text{eV}} \\ =656.08\text{nm} \end{gathered}$$

For the transition, $n=3\text{to}n=2$, the wavelength of photon is-

localid="1660057011319" $$\begin{gathered} {\lambda }_3=\frac{1240\text{eV}.\text{nm}}{-3.4\text{eV}-(-13.6)\text{eV}} \\ =\frac{1240\text{eV}·\text{nm}}{10.2\text{eV}} \\ =121.57\text{nm} \end{gathered}$$

The wavelengths of photon emitted during the three transitions- $n=3\text{to}n=1,n=2\text{to}n=1,\text{and}n=3\text{to}n=2$ are $102.56\text{nm,}656.08\text{nm},\text{and}121.57\text{nm}$ respectively.