Q 12-12-103JEE MainJEE Main 2020 (9 Jan, Shift 2)Medium
The energy required to ionise a hydrogen like ion in its ground state is $9$ Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?
Answer: (B) $11.4$ nm
Ionisation energy $= Z^2$ Rydberg $= 9$ Ry, so $Z = 3$.
Second excited state is $n = 3$:
$$\Delta E = Z^2(13.6)\left(1 - \frac{1}{9}\right) = 9\times13.6\times\frac{8}{9} = 108.8\ \text{eV}$$
$$\lambda = \frac{1240}{108.8}\ \text{nm} \approx 11.4\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics