Chapter 36, Problem 38P

(a)

To determine

The probability of finding the electron in the region between $r$ and $r+\Delta r$ .

Answer to Problem 38P

The probability of finding the electron in the region between $r$ and $r+\Delta r$ is $0.0162$ .

Explanation of Solution

Given:

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

The value of $\Delta r$ is $0.3a_0$ .

Formula used:

The expression for the normalized ground state wave function for an atom with number $Z$ is given by,

  $\psi \left(r\right)=\frac{1}{\sqrt{\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}{e}^{-\frac{Zr}{a_0}}$

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

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

The expression probability of finding the particle with region $\Delta r$ is given by,

  $P=P\left(r\right)\Delta r$

Calculation:

The atomic number of hydrogen atom is $1$ .

The normalized ground state wave function is calculated as,

  $\begin{array}{c}\psi \left(r\right)=\frac{1}{\sqrt{\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}{e}^{-\frac{\left(1\right)\left(a_0\right)}{a_0}}\\ =\frac{1}{\sqrt{\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}{e}^{-1}\\ =\frac{1}{e{a}_0\sqrt{\pi a_0}}\\ {\psi }^2\left(a_0\right)=\frac{1}{e^2{a}_0^3\pi }\end{array}$

The radial probability density is calculated as,

  $\begin{array}{c}P\left(r\right)=4\pi r^2{\left|\psi \left(r\right)\right|}^2\\ =4\pi {\left(a_0\right)}^2\left(\frac{1}{e^2{a}_0^3\pi }\right)\\ =\frac{4}{e^2{a}_0}\end{array}$

The probability of finding the particle with region $\Delta r$ is calculated as,

  $\begin{array}{c}P=P\left(r\right)\Delta r\\ =\frac{4}{e^2{a}_0}\left(0.03a_0\right)\\ =\frac{4}{{\left(2.718\right)}^2}\left(0.03\right)\\ =0.0162\end{array}$

Conclusion:

Therefore, the probability of finding the electron in the region between $r$ and $r+\Delta r$ is $0.0162$ .

(b)

To determine

The probability of finding the electron in the region between $r$ and $r+\Delta r$ .

Answer to Problem 38P

The probability of finding the electron in the region between $r$ and $r+\Delta r$ is $0.00879$ .

Explanation of Solution

Given:

The radius of Bohr bolt is $r=2a_0$ .

The value of $\Delta r$ is $0.3a_0$ .

Formula used:

The expression for the normalized ground state wave function for an atom with number $Z$ is given by,

  $\psi \left(r\right)=\frac{1}{\sqrt{\pi }}{\left(\frac{Z}{a_0}\right)}^{3/2}{e}^{-\frac{Zr}{a_0}}$

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

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

The expression probability of finding the particle with region $\Delta r$ is given by,

  $P=P\left(r\right)\Delta r$

Calculation:

The atomic number of hydrogen atom is $1$ .

The normalized ground state wave function is calculated as,

  $\begin{array}{c}\psi \left(r\right)=\frac{1}{\sqrt{\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}{e}^{-\frac{\left(1\right)\left(2a_0\right)}{a_0}}\\ =\frac{1}{\sqrt{\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}{e}^{-2}\\ =\frac{1}{e^2{a}_0\sqrt{\pi a_0}}\\ {\psi }^2\left(a_0\right)=\frac{1}{e^4{a}_0^3\pi }\end{array}$

The radial probability density is calculated as,

  $\begin{array}{c}P\left(2a_0\right)=4\pi r^2{\left|\psi \left(r\right)\right|}^2\\ =4\pi {\left(2a_0\right)}^2\left(\frac{1}{e^4{a}_0^3\pi }\right)\\ =\frac{16}{e^4{a}_0}\end{array}$

The probability of finding the particle with region $\Delta r$ is calculated as,

  $\begin{array}{c}P=P\left(r\right)\Delta r\\ =\frac{16}{e^4{a}_0}\left(0.03a_0\right)\\ =\frac{16}{{\left(2.718\right)}^4}\left(0.03\right)\\ =0.00879\end{array}$

Conclusion:

Therefore, the probability of finding the electron in the region between $r$ and $r+\Delta r$ is $0.00879$ .