GATE 2026 PH – Question 29
Consider an electron in the energy eigenstate $\psi_{211}(\mathbf r)$ of the hydrogen atom. Given that the radial probability distribution of the electron in such a state takes its maximum value at $r=n_0a$, where $a$ is the Bohr radius, and $n_0$ is an integer. The value of $n_0$ (in integer) is _____. The radial part of the wavefunction $\psi_{211}(\mathbf r)$ is given by $R_{21}(r)=\frac{r e^{-r/(2a)}}{\sqrt{24a^5}}$.
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Correct answer: 4
Explanation
The radial probability density is $P(r)=r^2|R_{21}(r)|^2$ (the factor $r^2$ comes from the volume element).
With $R_{21}=\dfrac{r\,e^{-r/2a}}{\sqrt{24a^5}}$:
$$P(r)\propto r^2\cdot r^2e^{-r/a}=r^4e^{-r/a}.$$
**Maximum:** set the derivative of $\ln P=4\ln r-r/a$ to zero:
$$\frac4r-\frac1a=0\;\Rightarrow\;r=4a .$$
So $n_0=\mathbf{4}$. (This is the radius of the $2p$ orbital's most probable electron position, $n^2a=4a$.)