Q 12-12-063JEE MainJEE Main 2023 (31 Jan, Shift 2)Easy
If the binding energy of ground state electron in a hydrogen atom is $13.6\ \text{eV}$, then the energy required to remove the electron from the second excited state of $\text{Li}^{2+}$ will be $x\times10^{-1}\ \text{eV}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 136
Second excited state is $n=3$. For $\text{Li}^{2+}$ ($Z=3$):
$$E=\frac{13.6Z^2}{n^2}=\frac{13.6\times9}{9}=13.6\ \text{eV}=136\times10^{-1}\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics