Question 39.41: What is the probability that an electron in the ground state......
What is the probability that an electron in the ground state of the hydrogen atom will be found between two spherical shells whose radii are r and r + Δr, (a) if r = 0.500a and
Δr = 0.010a and (b) if r = 1.00a and Δr = 0.01a, where a is the Bohr radius? (Hint: Δr is small enough to permit the radial probability density to be taken to be constant between r and r + Δr.)
Since Δr is small, we may calculate the probability using p = P(r)Δr, where P(r) is the radial probability density. The radial probability density for the ground state of hydrogen is given by Eq. 39-44:
P(r)=\frac{4}{a^3} r^2 e^{-2 r / a} (radial probability density, hydrogen atom ground state). (39-44)
where a is the Bohr radius.
(a) Here, r = 0.500a and Δr = 0.010a. Then,
P=\left(\frac{4 r^2 \Delta r}{a^3}\right) e^{-2 r / a}=4(0.500)^2(0.010) e^{-1}=3.68 \times 10^{-3} \approx 3.7 \times 10^{-3} .
(b) We set r = 1.00a and Δr = 0.010a. Then,
P=\left(\frac{4 r^2 \Delta r}{a^3}\right) e^{-2 r / a}=4(1.00)^2(0.010) e^{-2}=5.41 \times 10^{-3} \approx 5.4 \times 10^{-3} .