Question

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n=2 to n=1 state is:

• A. 121.8 nm
• B. 194.8 nm
• C. 490.7 nm
• D. 913.3 nm

Hint:

The transition from n=2 to n=1 in hydrogen is the first line of the Lyman series and is also known as the Lyman-alpha line. Its wavelength is a fundamental constant in atomic physics and astronomy, approximately 121.6 nm. Memorizing this can be helpful.

Answer 1

The Correct Option is A

To solve for the wavelength of the photon emitted when an electron transitions from the \(n = 2\) level to the \(n = 1\) level in a hydrogen atom, we use the Rydberg formula for hydrogen:

\(\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)

where:

• \(\lambda\) is the wavelength of the emitted photon.
• \(R_H \approx 1.097 \times 10^7 \, \text{m}^{-1}\) is the Rydberg constant for hydrogen.
• \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final states (with \(n_2 > n_1\)).

For this specific transition, \(n_1 = 1\) and \(n_2 = 2\). Plugging these values into the Rydberg formula, we have:

\(\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)\)

\(= 1.097 \times 10^7 \left( 1 - \frac{1}{4} \right)\)

\(= 1.097 \times 10^7 \times \frac{3}{4}\)

\(= 8.228 \times 10^6 \, \text{m}^{-1}\) 

Now, calculate the wavelength \(\lambda\):

\(\lambda = \frac{1}{8.228 \times 10^6} \approx 1.216 \times 10^{-7} \, \text{m}\)

Converting the wavelength from meters to nanometers (\(1 \, \text{m} = 10^9 \, \text{nm}\)):

\(\lambda \approx 121.6 \, \text{nm}\)

Thus, the wavelength of the photon emitted is approximately 121.8 nm, which matches the given correct answer.

Therefore, the correct option is 121.8 nm.