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
$\dfrac{1}{\lambda} = R\left(1 - \dfrac{1}{9}\right) = \dfrac{8R}{9} \approx 9.78 \times 10^6\ \text{m}^{-1}$, so $\lambda \approx 102$ nm.
Solution by Sreeraj P, M.Sc Physics