Chapter 26, Problem 94QAP

To determine

(a)

The radius of the first Bohr orbit of the hydrogen atom considering that the electron is bound to the proton by gravitational force (rather than electrostatic force)

Answer to Problem 94QAP

The radius of the first Bohr orbit of the hydrogen atom is given by the following expression $\text{r=}\frac{h^2}{4{\pi }^{2}G{m}_e{}^2{m}_p}$

Explanation of Solution

Given:

It is assumed that the electron is bound to the proton by gravitational force (rather than electrostatic force) in hydrogen atom.

Formula used:

In this case the centripetal force on the electron is equal to the gravitational force of attraction between the proton and electron. Therefore we can write

  $\frac{m_e{v}^2}{r}=\frac{G{m}_e{m}_p}{r^2}$

where m_e and m_p are electron and proton mass respectively, G is the gravitational constant.

According to Bohr quantization principle angular momentum of an electron in an atom is an integral multiple of h/2p.

So mathematically we can write it as

  $L=m_{e}vr=\frac{nh}{2\pi }$

where, m_e is the mass of the electron, v and r are the velocity of the electron and radius of the electron orbit respectively, n is a positive integer.

Calculation:

From Eq. (1.1) we can write

  $v^{2}r=G{m}_p$

Putting n =1 for the 1^st Bohr orbit in Eq. (1.2) we get

  $\begin{array}{l}{m}_{e}vr=\frac{h}{2\pi }\\ or,v\text{ =}\frac{h}{2\pi m_{e}r}\end{array}$

Using the expression for v in Eq. (1.3) we get

  $\begin{array}{l}\frac{h^2}{4{\pi }^2{m}_e{}^2{r}^2}r=G{m}_p\\ or,r\text{ =}\frac{h^2}{4{\pi }^{2}G{m}_e{}^2{m}_p}\end{array}$

Conclusion:

Therefore, the radius of the first Bohr orbit of the hydrogen atom is given by the following expression $\text{r =}\frac{h^2}{4{\pi }^{2}G{m}_e{}^2{m}_p}$

To determine

(b)

The energy of the electron in the first Bohr orbit

Answer to Problem 94QAP

The energy expression for the electron in the 1^st Bohr orbit is given by $E=-\frac{2{\pi }^2{G}^2{m}_e{}^3{m}_p{}^2}{h^2}$

Explanation of Solution

Given:

It is assumed that the electron is bound to the proton by gravitational force (rather than electrostatic force) in hydrogen atom.

Calculation:

Now substituting the expression for the radius of the 1^st Bohr orbit r given by Eq. 1.5 in Eq. 1.4 we get the expression for the velocity of the electron as

  $v=\frac{2\pi G{m}_e{m}_p}{h}$

So the kinetic energy of the electron is

  $E_k=\frac{1}{2}{m}_e{\left(\frac{2\pi G{m}_e{m}_p}{h}\right)}^2=\frac{2{\pi }^2{G}^2{m}_e{}^3{m}_p{}^2}{h^2}$

The gravitational potential energy of the electron is given by

  $E_p=-\frac{G{m}_{e}m}{r}$

Now substituting the expression for r from Eq. 1.5 in Eq. 1.8 we get

  $E_p=-\frac{4{\pi }^2{G}^2{m}_e{}^3{m}_p{}^2}{h^2}$

Therefore the total energy of the electron in the 1^st Bohr orbit is given by

  $E=E_k+E_p$

Now using the expressions for E_k (Eq. 1.7) and E_p (Eq. 1.9) we get the total energy as

  $E=-\frac{2{\pi }^2{G}^2{m}_e{}^3{m}_p{}^2}{h^2}$

Conclusion:

Therefore, the energy expression for the electron in the 1^st Bohr orbit is given by $E=-\frac{2{\pi }^2{G}^2{m}_e{}^3{m}_p{}^2}{h^2}$