Question

An excited hydrogen atom emits light with a wavelength of \(397.2 \mathrm{~nm}\) to reach the energy level for which \(n=2\). In which principal quantum level did the electron begin?

Short answer

The electron initially started in the principal quantum level \(n_{2} = 3\).

Step-by-step solution

Recall the Rydberg formula

The Rydberg formula is given by: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_{1}^2} - \frac{1}{n_{2}^2}\right)\] where \(\lambda\) is the wavelength of emitted light, \(R_H\) is the Rydberg constant for hydrogen \((R_H = 1.097 \times 10^7 \, \text{m}^{-1})\), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.

Insert known values into the Rydberg formula

We know the final energy level \(n_1 = 2\), and we are given the wavelength of the emitted light \(\lambda = 397.2\, \text{nm}\). We need to find the initial energy level \(n_2\). Convert the given wavelength from nanometers to meters: \[\lambda = 397.2\, \text{nm} \times \frac{1\, \text{m}}{10^9\, \text{nm}} = 397.2 \times 10^{-9}\, \text{m}\] Now, we insert the known values into the Rydberg formula and solve for \(n_2\): \[\frac{1}{397.2\times10^{-9} \, \text{m}} = R_H \left(\frac{1}{4} - \frac{1}{n_{2}^2}\right)\]

Solve for the initial energy level, n2

Rearrange the equation and solve for \(n_{2}^2\): \[n_{2}^2 = \frac{1}{\frac{1}{R_H\lambda}+\frac{1}{4}}\]

Calculate n2, the initial energy level

Substitute the given values into the equation: \[n_{2}^2 = \frac{1}{\frac{1}{(1.097\times10^7 \, \text{m}^{-1})(397.2\times10^{-9} \, \text{m})}+\frac{1}{4}}\] Calculate \(n_{2}^2\): \[n_{2}^2 \approx 6.003\] Since the principal quantum level must be a positive integer, we round up to the nearest whole number: \[n_{2} = 3\] So, the electron initially started in the principal quantum level \(n_{2} = 3\).