Chapter 26, Problem 77QAP

To determine

(a)

The angular momentum of an electron in the n^thstate of the hydrogen atom

Answer to Problem 77QAP

Mathematically the angular momentum of an electron can be written as
  $L=m{v}_n{r}_n=\frac{nh}{2\pi }$

Explanation of Solution

Given:

Hydrogen atom model according to Bohr quantization principle

Calculation:

According to Bohr quantization principle angular momentum of an electron in an atom is an integral multiple of $\frac{h}{2\pi }$.

So mathematically we can write it as,
  $L=m{v}_n{r}_n=\frac{nh}{2\pi }$
(1)
where, m is the mass of the electron, v_nis the velocity on the n^thorbit, n is a positive integer.

Conclusion:

So mathematically the angular momentum of an electron can be written as
  $L=m{v}_n{r}_n=\frac{nh}{2\pi }$

To determine

(b)

The radius of the electron orbit in hydrogen atom

Answer to Problem 77QAP

The radius of the electron orbit in hydrogen atom is given by $r_n=\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}$

Explanation of Solution

Given:

Hydrogen atom model according to Bohr quantization principle

Calculation:

Electrostatic force between the orbiting electron and proton supplies the required centripetal force. So equating these two forces we get
$\frac{m{v}_n{}^2}{r_n}=\frac{k{e}^2}{r_n{}^2}$
(2)
where $(k=1/4\pi {\epsilon }_0)$ and for hydrogen atom we have assumed Z = 1 (1 proton at the nucleus). So, equation (2) can be written as

  $m{v}_n{}^2{r}_n=k{e}^2$
(3)
Now using equation (1) in equation (3) we can write,

  $\begin{array}{l}{(}^m\frac{1}{m{r}_n}=k{e}^2\\ \text{or, }{\left(\frac{nh}{2\pi }\right)}^2\frac{1}{m{r}_n}=k{e}^2\\ \text{or, }{r}_n=\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}\end{array}$
(4)

Conclusion:

So, the radius of the n^thelectron orbit in hydrogen atom is given by $r_n=\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}$

To determine

(c)

The kinetic energy of an electron in hydrogen atom

Answer to Problem 77QAP

The kinetic energy of an electron in hydrogen atom is given by $K_n=\frac{m{k}^2{e}^4}{2n^2{\hslash }^2}$

Explanation of Solution

Given:

Hydrogen atom model according to Bohr quantization principle

Calculation:

Kinetic energy of the electron is given by
  $K_n=\frac{1}{2}m{v}_n{}^2$
(5)

Now substituting the expression for r_nin equation (1) we obtain
  $\begin{array}{l}m{v}_n\left(\frac{n^2{h}^2}{4{\pi }^{2}mk{e}^2}\right)=\frac{nh}{2\pi }\\ \text{or, }{v}_n=\left(\frac{4{\pi }^{2}mk{e}^{2}nh}{2\pi n^2{h}^{2}m}\right)=\frac{2\pi k{e}^2}{nh}\end{array}$
(6)

Now substituting the value of v_nin equation (5) we obtain

  $K_n=\frac{1}{2}m{\left(\frac{2\pi k{e}^2}{nh}\right)}^2=\frac{m{k}^2{e}^4}{2n^2{\hslash }^2}$
(7)

Conclusion:

So, the kinetic energy of an electron in hydrogen atom is given by $K_n=\frac{m{k}^2{e}^4}{2n^2{\hslash }^2}$

To determine

(d)

The total energy of an electron in hydrogen atom

Answer to Problem 77QAP

The total energy of an electron in hydrogen atom is given by $E_n\text{=}-\frac{m{k}^2{e}^4}{2n^2{\hslash }^2}$

Explanation of Solution

Given:

Hydrogen atom model according to Bohr quantization principle

Calculation:

Total energy of the electron is given by the sum of kinetic and potential energy
  $E_n=K_n+K_p$
(8)

K_nexpression we have already derived in part (c). Now we want to derive an expression for the potential energy E_pwhich is the electro static energy between proton and the orbiting electron. So
$$

  $K_p=-\frac{k{e}^2}{r_n}$
(9)
Now substituting the value of r_nusing equation (4) in equation (9) we get

  $K_p=-k{e}^2\left(\frac{4{\pi }^{2}mk{e}^2}{n^2{h}^2}\right)=-\frac{m{k}^2{e}^4}{n^2{\hslash }^2}$
(10)
Now putting equation (7) and (10) in equation (8) we get

  $\begin{array}{l}{E}_n=\frac{m{k}^2{e}^4}{2n^2{\hslash }^2}-\frac{m{k}^2{e}^4}{n^2{\hslash }^2}\\ \text{or, }{E}_n\text{=}-\frac{m{k}^2{e}^4}{2n^2{\hslash }^2}\end{array}$

Conclusion:

So, the total energy of an electron in hydrogen atom is given by $E_n\text{=}-\frac{m{k}^2{e}^4}{2n^2{\hslash }^2}$

To determine

(e)

The speed of an electron in hydrogen atom

Answer to Problem 77QAP

The speed of an electron in hydrogen atom is given by $v_n=\frac{2\pi k{e}^2}{nh}$

Explanation of Solution

Given:

Hydrogen atom model according to Bohr quantization principle

Calculation:

We have already derived the expression for speed of an electron in part (c) equation (6). The expression for speed in the nth orbit of the electron is given by

  $v_n=\left(\frac{4{\pi }^{2}mk{e}^{2}nh}{2\pi n^2{h}^{2}m}\right)=\frac{2\pi k{e}^2}{nh}$

Conclusion:

So, the speed of an electron in hydrogen atom is given by $v_n=\frac{2\pi k{e}^2}{nh}$