Chapter 39: Q46P (page 1217)

Calculate the probability that the electron in the hydrogen atom, in its ground state, will be found between spherical shells whose radii are a and 2a , where a is the Bohr radius?

Short Answer

Expert verified

The required probability is P=0.439.

Step by step solution

Identification of the given data:

The given data is listed below.

Radii of the spherical shells are given as a and 2a .

Formula for finding the probability of electron:

The ground state wave function of hydrogen atom is given by,

$ψ_{100} (r, θ, ϕ) = \frac{1}{\sqrt{π}} {(\frac{1}{a})}^{\frac{3}{2}} e^{- \frac{r}{a}}$

Here, the Bohr radius is $a = 5.292 \times 10^{- 11} m$ .

Determine the probability of the electron in the hydrogen atom in its ground state:

The probability of finding the electron found between spherical shells is,

$P = \int_{a}^{2 a} \int_{0}^{π} \int_{0}^{2 π} {|ψ_{100} (r, θ, ϕ)|}^{2} r^{2} drsinθdθdϕ = \int_{a}^{2 a} {|\frac{1}{\sqrt{π}} {(\frac{1}{a})}^{\frac{3}{2}} e^{\frac{- r}{a}}|}^{2} r^{2} dr \int_{0}^{π} dϕ = \frac{1}{π} {(\frac{1}{a})}^{3} \int_{a}^{2 a} {|e^{- \frac{r}{a}}|}^{2} r^{2} dr \times {(- cosθ)}_{0}^{π} {(θ)}_{0}^{2 π} = \frac{1}{π} {(\frac{1}{a})}^{3} {[- \frac{1}{4} {ae}^{\frac{- 2 r}{a}} (a^{2} + 2 ar + 2 r^{2})]}_{a}^{2 a} \times (- cosπ + \cos 0) (2 π - 0)$

$P = \frac{1}{π} {(\frac{1}{a})}^{3} [- \frac{a}{4} \{e^{\frac{- 2 (2 a)}{a}} (a^{2} + 2 a (2 a) + 2 {(2 a)}^{2}) - e^{\frac{- 2 (a)}{a}} (a^{2} + 2 a^{2} + 2 a^{2})\}] \times (- (- 1) + 1) (2 π) = \frac{1}{π} {(\frac{1}{a})}^{3} [- \frac{a}{4} \{e^{- 4} (13 a^{2}) - e^{- 2} (5 a^{2})\}] \times (4 π) = 4 {(\frac{1}{a})}^{3} [- \frac{a}{4} \{0.0183 (13 a^{2}) - 0.1353 (5 a^{2})\}] P = 4 [- \frac{1}{4} \{0.0183 (13) - 0.1353 (5)\}] = - 4 [\frac{1}{4} \{0.2379 - 0.6765\}] = 0.439$