Answer
$=5.41\times 10^{-3}$
Work Step by Step
The probability of finding an electron in the ground state of the hydrogen atom between two spherical shells whose radii are $r$ and $r+\Delta r$ is
$\;\;\;\;\int_r^{r+\Delta r} P(r)dr$
Here, $r$ is small enough to permit the radial probability density to be taken to be constant between $r$ and $r+\Delta r$
Therefore,
$\;\;\;\;\int_r^{r+\Delta r} P(r)dr=P(r)\Delta r=\frac{4}{a^3}r^2e^{-2r/a}\Delta r$
Putting known values, we obtain,
$\frac{4}{a^3}r^2e^{-2r/a}\Delta r\\
=\frac{4}{a^3}\times(1.00a)^2\times e^{-2\times1.00a/a}\times 0.01a\\
=4\times1.00^2\times e^{-2}\times 0.01\\
=5.41\times 10^{-3}$
