Q 12-12-114JEE MainJEE Main 2019 (11 Jan, Shift 2)Easy
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
Answer: (D) $\dfrac{20}{27}\lambda$
M, L, N are $n = 3, 2, 4$.
$$\frac1\lambda \propto \frac14 - \frac19 = \frac5{36},\qquad \frac1{\lambda'} \propto \frac14 - \frac1{16} = \frac3{16}$$
$$\lambda' = \lambda\times\frac{5/36}{3/16} = \frac{20}{27}\lambda$$
Solution by Sreeraj P, M.Sc Physics