Q 12-12-109JEE MainJEE Main 2019 (10 Apr, Shift 2)Medium
In $\text{Li}^{++}$, an electron in the first Bohr orbit is excited to a level by a radiation of wavelength $\lambda$. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $\lambda$? (Given: $h = 6.63\times10^{-34}\ \text{J s}$; $c = 3\times10^8\ \text{m s}^{-1}$)
Answer: (A) $10.8\ \text{nm}$
Six lines means $\dfrac{n(n-1)}{2} = 6$, so the electron was excited to $n = 4$.
For $\text{Li}^{++}$, $Z = 3$:
$$\Delta E = 13.6\times9\left(1 - \frac1{16}\right) = 114.75\ \text{eV}$$
$$\lambda = \frac{hc}{\Delta E} = \frac{1240\ \text{eV nm}}{114.75\ \text{eV}} \approx 10.8\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics