Answer
See the detailed answer below.
Work Step by Step
We know that the radial wave function for the hydrogen atom in the 1s state is given by:
$$
R_{1s}(r) = A e^{-r/a_B}
$$
where $ A $ is the normalization constant, and $ a_B $ is the Bohr radius.
The radial wave function must satisfy the normalization condition for the three-dimensional hydrogen atom
$$
4\pi \int_0^\infty |R_{1s}(r)|^2 r^2 \, dr = 1
$$
Substituting $ R_{1s}(r) = A e^{-r/a_B} $:
$$
4\pi\int_0^\infty \left( A e^{-r/a_B} \right)^2 r^2 \, dr = 1
$$
$$
4\pi A^2 \int_0^\infty e^{-2r/a_B} r^2 \, dr = 1\tag 1
$$
We can use the provided integral formula:
$$
\int_0^\infty x^n e^{-\alpha x} \, dx = \frac{n!}{\alpha^{n+1}}
$$
where, in our case, $ x = r $, $ n = 2 $, $ \alpha = 2/a_B $.
Thus, the integral becomes:
$$
\int_0^\infty e^{-2r/a_B} r^2 \, dr = \frac{2!}{(2/a_B)^3} = \frac{2}{(2/a_B)^3} = \frac{a_B^3}{4}
$$
Substitute this result into (1):
$$
4\pi A^2\cdot \frac{a_B^3}{4} = 1
$$
Solving for $ A^2 $:
$$
A^2 = \frac{\color{red}{\bf\not} 4}{\color{red}{\bf\not} 4\pi a_B^3}
$$
Taking the square root:
$$
A = \frac{1}{\pi a_B^{3/2}}
$$
Therefore,
$$
\boxed{ A = \left(\pi a_B^3\right)^{-1/2}}
$$
