Chapter 28, Problem 25E

(a)

To determine

The ratio of the radial probabilities at $r_t$ and at $a_0$.

Answer to Problem 25E

The ratio of the radial probabilities at $r_t$ and at $a_0$ is $\underset{\_}{4}$.

Explanation of Solution

In a spherical shell, the probability of finding an electron is directly proportional to the radial probability.

Write the expression for the radial probability,

  $r^{2}P=r^2{\left|\psi \right|}^2$

Here, the radius is $r$, probability is $P$, and $\psi$ is the wave function.

Use $r_t$ for $r$.

  $r_t^{2}P=r_t^2{\left|\psi \right|}^2$        (I)

Write the expression for the classical turning point.

  $r_t=2n^2{a}_0$        (II)

Here, the classical turning point is $r_t$, principal quantum number is $n$, and $a_0$ is Bohr radius.

Use equation (II) in equation (I).

  $r_t^{2}P={\left(2n^2{a}_0\right)}^2{\left|\psi \right|}^2$

Here, $n=1$, so the above equation becomes,

  $r_t^{2}P={\left(2a_0\right)}^2{\left|\psi \right|}^2$

  $r_t^{2}P=4a_0^2{\left|\psi \right|}^2$        (III)

The radial probability at $r_t$ is $r_t^{2}P=4a_0^2{\left|\psi \right|}^2$.

For $r=a_0$,

Use $a_0$ for $r$.

  $a_0^{2}P=a_0^2{\left|\psi \right|}^2$        (IV)

The radial probability for $a_0$ is $a_0^{2}P=a_0^2{\left|\psi \right|}^2$.

Conclusion:

Divide equation (III) by (IV).

  $\begin{array}{c}\frac{r_t^{2}P}{a_0^{2}P}=\frac{4a_0^2{\left|\psi \right|}^2}{a_0^2{\left|\psi \right|}^2}\\ \frac{r_t^{2}P}{a_0^{2}P}=4\\ \frac{r_t^2}{a_0^2}=4\end{array}$

Therefore, the ratio of the radial probabilities at $r_t$ and at $a_0$ is $\underset{\_}{4}$.

(b)

To determine

The ratio of the radial probabilities at $2r_t$ and at $a_0$.

Answer to Problem 25E

The ratio of the radial probabilities at $2r_t$ and at $a_0$ is $\underset{\_}{16}$.

Explanation of Solution

In a spherical shell, the probability of finding an electron is directly proportional to the radial probability.

Write the expression for the radial probability,

  $r^{2}P=r^2{\left|\psi \right|}^2$

Here, the radius is $r$, probability is $P$, and $\psi$ is the wave function.

Use $2r_t$ for $r$.

  ${\left(2r_t\right)}^{2}P={\left(2r_t\right)}^2{\left|\psi \right|}^2$        (V)

Write the expression for the classical turning point.

  $r_t=2n^2{a}_0$        (VI)

Here, the classical turning point is $r_t$, principal quantum number is $n$, and $a_0$ is Bohr radius.

  $2r_t=2\left(2n^2{a}_0\right)$

Use equation (II) in equation (I).

  ${\left(2r_t\right)}^{2}P={\left(4n^2{a}_0\right)}^2{\left|\psi \right|}^2$

Here, $n=1$, so the above equation becomes,

  ${\left(2r_t\right)}^{2}P={\left(4a_0\right)}^2{\left|\psi \right|}^2$

  ${\left(2r_t\right)}^{2}P={\left(4a_0\right)}^2{\left|\psi \right|}^2$        (VII)

The radial probability at $2r_t$ is ${\left(2r_t\right)}^{2}P={\left(4a_0\right)}^2{\left|\psi \right|}^2$.

For $r=a_0$,

Use $a_0$ for $r$.

  $a_0^{2}P=a_0^2{\left|\psi \right|}^2$        (VIII)

The radial probability for $a_0$ is $a_0^{2}P=a_0^2{\left|\psi \right|}^2$.

Conclusion:

Divide equation (VII) by (VIII).

  $\begin{array}{c}\frac{{\left(2r_t\right)}^{2}P}{a_0^{2}P}=\frac{{\left(4a_0\right)}^2{\left|\psi \right|}^2}{a_0^2{\left|\psi \right|}^2}\\ \frac{{\left(2r_t\right)}^{2}P}{a_0^{2}P}=16\\ \frac{{\left(2r_t\right)}^2}{a_0^2}=16\end{array}$

Therefore, the ratio of the radial probabilities at $2r_t$ and at $a_0$ is $\underset{\_}{16}$.