Answer
See the detailed answer below.
Work Step by Step
To determine the most probable distance between the proton and the electron in the 2s state of hydrogen, we need to analyze the radial probability density $ P_r(r) $ for this state.
The radial probability density $ P_r(r) $ for the 2s state is given by
$$
P_r(r) = 4\pi r^2 |R_{2s}(r)|^2\tag 1
$$
where $R_{2s}(r)$ is given by
$$
R_{2s}(r) = $$
Plug into (1);
$$
P_r(r) = 4\pi r^2 \left[\frac{1}{\sqrt{8\pi a_B^3}} \left(1 - \frac{r}{2a_B}\right) e^{-r/2a_B}\right]^2
$$
$$
P_r(r) = \dfrac{ 1}{ 2 a_B^3} \left(1 - \frac{r}{2a_B}\right)^2r^2 e^{-r/a_B}
$$
To find the most probable distance, we need to locate the maximum of the radial probability density function. This requires taking the derivative of $ P_r(r) $ with respect to $ r $ and setting it equal to zero.
$$
\frac{dP_r}{dr} =\frac{d}{dr} \left[\dfrac{ 1}{ 2 a_B^3} \left(1 - \frac{r}{2a_B}\right)^2r^2 e^{-r/a_B} \right]=0
$$
where $\dfrac{ 1}{ 2 a_B^3}$ is constant.
$$ \frac{d}{dr} \left[ \left(1 - \frac{ r}{ a_B}+\frac{r^2}{4a_B^2}\right) r^2 e^{-r/a_B} \right]=0 $$
$$ \frac{d}{dr} \left[ \left(r^2 - \frac{ r^3}{ a_B }+\frac{r^4}{4a_B^2}\right) e^{-r/a_B} \right]=0 $$
$$= \left(2r - \frac{ 3r^2}{ a_B }+\frac{r^3}{ a_B^2}\right) e^{-r/a_B} +\\
\left(r^2 - \frac{ r^3}{ a_B }+\frac{r^4}{4a_B^2}\right) \left[\dfrac{-1}{a_B}\; e^{-r/a_B} \right]=0 $$
Since both terms include the exponential factor $ e^{-r/a_B}$, and this function is never zero, we can eliminate the exponential term:
$$
\left( 2r - \frac{3r^2}{a_B} + \frac{r^3}{ a_B^2} \right) - \frac{1}{a_B} \left( r^2 - \frac{r^3}{a_B} + \frac{r^4}{4a_B^2} \right) = 0
$$
$$ 2r - \frac{3r^2}{a_B} + \frac{r^3}{ a_B^2} - \frac{r^2}{a_B} + \frac{r^3}{a_B^2} - \frac{r^4}{4a_B^3} = 0
$$
$$ 2r - \frac{4r^2}{a_B} + \frac{2r^3}{ a_B^2} - \frac{r^4}{4a_B^3} = 0
$$
$$ r \left[ 2 - \frac{4r }{a_B} + \frac{2r^2}{ a_B^2} - \frac{r^3}{4a_B^3} \right] = 0
$$
we can see that $r=0$ which is dismissed here and not reasonable.
$$ 2 - \frac{4r }{a_B} + \frac{2r^2}{ a_B^2} - \frac{r^3}{4a_B^3} = 0$$
This cubic equation has three roots, reflecting the fact that the radial probability density can have multiple extrema two maxima, and one minimum. We now focus on identifying the specific root that corresponds to the most probable distance.
In this case, we only need to verify that $ r = 5.236 a_B $ satisfies the equation.
Substituting $ r = 5.236 a_B $ into the cubic equation:
$$2 - \frac{4(5.236 a_B) }{a_B} + \frac{2(5.236 a_B)^2}{ a_B^2} - \frac{(5.236 a_B)^3}{4a_B^3} =x $$
If $x=0$, then that confirms that $ r = 5.236 a_B $ is indeed a solution.
$$
2 - 4(5.236) + 2(5.236 a_B)^2 - \frac{(5.236 a_B)^3}{4 } =x=0.00025
$$
$$x\approx 0$$
Thus, the most probable distance between the proton and the electron in the 2s state of hydrogen is $ r = 5.236 a_B $, where $ a_B $ is the Bohr radius.
