Q 12-12-077JEE MainJEE Main 2023 (6 Apr, Shift 2)Medium
Experimentally it is found that $12.8\ \text{eV}$ energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is $\dfrac9x\times10^{-10}$ m. The value of $x$ is ______. ($1\ \text{eV}=1.6\times10^{-19}$ J, $\dfrac1{4\pi\epsilon_0}=9\times10^9\ \text{N m}^2\,\text{C}^{-2}$ and electronic charge $=1.6\times10^{-19}$ C)
Numerical value type. Enter your answer.
Answer: 16
Ionisation energy $=\dfrac{ke^2}{2r}$:
$$r=\frac{9\times10^9\times(1.6\times10^{-19})^2}{2\times12.8\times1.6\times10^{-19}}=\frac{14.4\times10^{-10}}{25.6}=\frac{9}{16}\times10^{-10}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics