Q 12-12-028JEE MainJEE Main 2026 (5 Apr, Shift 1)Easy
In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit ($f$). The ratio of the radius of the orbits in given by $r_i : r_f = 16 : 4$. The wavelength of photon emitted due to this transition is ______ nm.
(Given Rydberg constant $= 1.0973 \times 10^7$/m)
Answer: (C) $486$
$r \propto n^2$, so $n_i = 4$ and $n_f = 2$.
$$\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16} \;\Rightarrow\; \lambda = \frac{16}{3 \times 1.0973 \times 10^7} \approx 486\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics