Answer
See the answer below.
Work Step by Step
a)
$\psi=\dfrac 1{\sqrt{\pi.(52.9\times10^{-12}\ m)^3}}.e^{-\frac{r=a_0}{a_0}}$
$\psi=5.39\times10^{14}\ m^{-3/2}$
$P=4\pi.(52.9\times10^{-12}\ m)^2\times(5.39\times10^{14}\ m^{-3/2})^2\times1.0\times10^{-12}\ m$
$P=0.01\approx 1\%$
b) Similarly as in a:
$0.50 a_0: P=6.95\times10^{-3}$
$4a_0: P=4.06\times10^{-4}$
The decline in probability away from the most probable distance is in accord with figure 6.12b.
