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) $486$
Concept. $r\propto n^2$, so the radius ratio $16:4$ gives $n_i=4$, $n_f=2$. Then $\dfrac1\lambda=R\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)$.
Step 1. Wave number. $\dfrac1\lambda=R\left(\dfrac14-\dfrac1{16}\right)=\dfrac{3R}{16}$.
Step 2. Wavelength. $\lambda=\dfrac{16}{3\times1.0973\times10^7}=4.86\times10^{-7}$ m $=486$ nm.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
