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

The wavelength of a photon emitted during an electron transition in a hydrogen atom is given by the Rydberg formula:
$\frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$.
Here, $R_H$ is the Rydberg constant, $n_i$ is the initial energy level, and $n_f$ is the final energy level.
For this transition, the initial state is $n_i = 2$ and the final state is $n_f = 1$.
The value of the Rydberg constant is $R\_H \approx 1.097 \times 10^7$ m⁻¹.
Substituting the values into the formula:
$\frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = (1.097 \times 10^7) \left(1 - \frac{1}{4}\right)$.
$\frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{3}{4}\right)$.
$\lambda = \frac{4}{3 \times 1.097 \times 10^7} \approx 1.215 \times 10^{-7}$ m.
To convert this wavelength to nanometers, we multiply by $10^9$:
$\lambda = 1.215 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = 121.5$ nm.
This value is very close to option (A), 121.8 nm. The small difference is due to using a more precise value of the Rydberg constant.