Answer
a. 0.0803.
b. 0.243.
Work Step by Step
a. The probability is $P=\int^{a/2}_{0}|\Psi_{1s}|^2 4\pi r^2dr$
The expression is evaluated in Example 41.4, and yields $1-\frac{5}{2}e^{-1}=0.0803$.
b. In Example 41.4, we calculated the probability for the electron to be less than a distance a from the nucleus; it is $1-5e^{-2}$.
Now we have just calculated the probability for the electron to be less than a distance a/2 from the nucleus. By subtracting, we find the probability of the electron being in the range of a/2 to a.
$(1-5e^{-2}) -(1-\frac{5}{2}e^{-1})=0.243$
