Q 12-12-020NEETJEE MainMedium
An electron in a hydrogen atom jumps from $n = 3$ to $n = 1$. The wavelength of the emitted photon is ($R = 1.1 \times 10^7\ \text{m}^{-1}$)
Answer: (A) $102$ nm
Concept. Rydberg formula: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$.
Step 1. Substitute. $n_1 = 1$, $n_2 = 3$:
$$\frac{1}{\lambda} = R\left(1 - \frac{1}{9}\right) = \frac{8 \times 1.1 \times 10^7}{9} \approx 9.78 \times 10^6\ \text{m}^{-1}$$
Step 2. Wavelength. $\lambda \approx 1.02 \times 10^{-7}$ m $= 102$ nm.
Answer. $102$ nm (Option A).
Solution by Sreeraj P, M.Sc Physics