Answer
See the detailed answer below.
Work Step by Step
In this problem, we need to find the radial wave function and the radial probability density for 3 different values of $ r $, using the hydrogen atom’s 1s state.
$$\color{blue}{\bf [a]}$$
$\underline{\text{For $ r = \frac{1}{2} a_B $}}$
$\bullet$ The radial wave function is given by
$$
R_{1s}\left(\frac{1}{2} a_B\right) = \frac{1}{\sqrt{\pi a_B^3}} e^{-\frac{1}{2}} = \frac{0.607}{\sqrt{\pi a_B^3}}
$$
$$
R_{1s}\left(\frac{1}{2} a_B\right) =\color{red}{\bf 0.342 }\, a_B^{-3/2}
$$
$\bullet$ $\bullet$ The radial probability density is:
$$
P_r\left(\frac{1}{2} a_B\right) = 4\pi \left(\frac{a_B}{2}\right)^2 \left( \frac{0.607}{\sqrt{\pi a_B^3}} \right)^2 = \frac{0.368}{a_B}
$$
$$
P_r\left(\frac{1}{2} a_B\right) =\color{red}{\bf 0.368}\;a_B^{-1}
$$
$\bullet$ $\bullet$ $\bullet$ The probability of finding the electron within a small shell $ \delta r = 0.010\; a_B $ at $ r = \frac{1}{2} a_B $ is:
$$
\text{Prob}(r) = P_r\left(\frac{1}{2} a_B\right) \delta r = \frac{0.368}{a_B} (0.010 a_B)
$$
$$
\text{Prob}(r) = \color{red}{\bf 3.7 \times 10^{-3}}
$$
$$\color{blue}{\bf [b]}$$
$\underline{\text{For $ r = a_B $}}$
$\bullet$ The radial wave function is:
$$
R_{1s}(a_B) = \frac{1}{\sqrt{\pi a_B^3}} e^{-1} = \frac{0.368}{\sqrt{\pi a_B^3}}
$$
$$
R_{1s}(a_B) =\color{red}{\bf 0.208}\, a_B^{-3/2}
$$
$\bullet$$\bullet$ The radial probability density is:
$$
P_r(a_B) = 4\pi a_B^2 \left(\frac{0.368}{\sqrt{\pi a_B^3}}\right)^2 = \frac{0.541}{a_B}
$$
$$
P_r(a_B) = \color{red}{\bf 0.541}\;a_B^{-1}
$$
$\bullet$$\bullet$$\bullet$The probability of finding the electron in a small shell $ \delta r = 0.010\; a_B $ at $ r = a_B $ is:
$$
\text{Prob}(r) = P_r(a_B) \delta r = \frac{0.541}{a_B} (0.010 \;a_B)
$$
$$
\text{Prob}(r) = \color{red}{\bf 5.41 \times 10^{-3}}
$$
$$\color{blue}{\bf [c]}$$
$\underline{\text{For $ r = 2a_B $}}$
$\bullet$ The radial wave function is:
$$
R_{1s}(2a_B) = \frac{1}{\sqrt{\pi a_B^3}} e^{-2} = \frac{0.135}{\sqrt{\pi a_B^3}}
$$
$$
R_{1s}(2a_B) = \color{red}{\bf 0.076}\, a_B^{-3/2}
$$
$\bullet$$\bullet$ The radial probability density is:
$$
P_r(2a_B) = 4\pi (2a_B)^2 \left(\frac{0.135}{\sqrt{\pi a_B^3}}\right)^2 = \frac{0.293}{a_B}
$$
$$
P_r(2a_B) = \color{red}{\bf 0.293}\;{a_B^{-1}}
$$
$\bullet$$\bullet$$\bullet$ The probability of finding the electron in a small shell $ \delta r = 0.010\; a_B $ at $ r = 2a_B $ is:
$$
\text{Prob}(r) = P_r(2a_B) \delta r = \frac{0.293}{a_B} (0.010\; a_B)
$$
$$
\text{Prob}(r)=\color{red}{\bf 2.9 \times 10^{-3}}
$$
