Chapter 36, Problem 39P

(a)

To determine

The value of $\psi \left(r\right)$ at $r=a_0$ .

Answer to Problem 39P

The value of $\psi \left(r\right)$ at $r=a_0$ is $\frac{1}{4\sqrt{2e\pi }{\left(a_0\right)}^{3/2}}$ .

Explanation of Solution

Given:

The radius of Bohr bolt is $r=a_0$ .

Formula used:

The expression for the spherical wave function for $n=2,l=0,m_{\text{l}}=0$ is given by,

  ${\psi }_{200}=C_{200}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}$

The expression for the constant term $C_{200}$ is given by,

  $C_{200}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}$

The new expression for the spherical wave function is given as,

  $\begin{array}{c}{\psi }_{200}=C_{200}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}\\ =\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}\end{array}$

Calculation:

The atomic number of hydrogen atom is $1$ .

The spherical wave function for $n=2,l=0,m_{\text{l}}=0$ is calculated as,

  $\begin{array}{c}{\psi }_{200}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}\\ =\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{1\left(a_0\right)}{a_0}\right)e^{-\left(1\right)\left(a_0\right)/2a_0}\\ =\frac{1}{4\sqrt{2e\pi }{\left(a_0\right)}^{3/2}}\end{array}$

Conclusion:

Therefore, the value of $\psi \left(r\right)$ at $r=a_0$ is $\frac{1}{4\sqrt{2e\pi }{\left(a_0\right)}^{3/2}}$ .

(b)

To determine

The value of ${\psi }^2\left(r\right)$ at $r=a_0$ .

Answer to Problem 39P

The value of ${\psi }^2\left(r\right)$ at $r=a_0$ is $\frac{0.00366}{a_0^3}$ .

Explanation of Solution

Given:

The radius of Bohr bolt is $r=a_0$ .

Formula used:

The expression for the spherical wave function for $n=2,l=0,m_{\text{l}}=0$ is given by,

  ${\psi }_{200}=C_{200}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}$

The expression for the constant term $C_{200}$ is given by,

  $C_{200}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}$

The new expression for the spherical wave function is given as,

  $\begin{array}{c}{\psi }_{200}=C_{200}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}\\ =\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}\end{array}$

Calculation:

The atomic number of hydrogen atom is $1$ .

The spherical wave function for $n=2,l=0,m_{\text{l}}=0$ is calculated as,

  $\begin{array}{c}{\psi }_{200}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}\left(2-\frac{Zr}{a_0}\right)e^{-Zr/2a_0}\\ =\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{1\left(a_0\right)}{a_0}\right)e^{-\left(1\right)\left(a_0\right)/2a_0}\\ {\psi }_{200}^2={\left(\frac{1}{4\sqrt{2e\pi }{\left(a_0\right)}^{3/2}}\right)}^2\\ {\psi }_{200}^2=\frac{0.00366}{a_0^3}\end{array}$

Conclusion:

Therefore, the value of ${\psi }^2\left(r\right)$ at $r=a_0$ is $\frac{0.00366}{a_0^3}$ .

(c)

To determine

The value of radial probability density $P\left(r\right)$ at $r=a_0$ .

Answer to Problem 39P

The value of radial probability density $P\left(r\right)$ at $r=a_0$ is $\frac{0.0460}{a_0}$ .

Explanation of Solution

Given:

The radius of Bohr bolt is $r=a_0$ .

Formula used:

The expression for the radial probability density of finding a particle at a position $r$ is given by,

  $P\left(r\right)=4\pi r^2{\left|\psi \left(r\right)\right|}^2$

Calculation:

The radial probability density is calculated as,

  $\begin{array}{c}P\left(r\right)=4\pi r^2{\left|\psi \left(r\right)\right|}^2\\ =4\pi {\left(a_0\right)}^2\left(\frac{0.00366}{a_0^3}\right)\\ =\frac{0.0460}{a_0}\end{array}$

Conclusion:

Therefore, the value of radial probability density $P\left(r\right)$ at $r=a_0$ is $\frac{0.0460}{a_0}$ .