Chapter 27, Problem 75GP

To determine

The ratio of the gravitational to the electrical force.

To identify: Whether the gravitational force can be safely ignored or not.

Answer to Problem 75GP

  $4.4\times {10}^{-40}$ . Yes, maybe the gravitational force is safely ignored.

Explanation of Solution

Given:

The hydrogen atom is given.

Formula used:

It is known that: $F_g=\left(\frac{G{m}_e{m}_p}{r^2}\right)and{F}_e=\left(\frac{k{e}^2}{r^2}\right)$

Where,

  $G=\left(\text{6}\text{.67}\times {\text{10}}^{-11}\right)$ is gravitational constant.

  $k$ is constant factor.

  $F_g$ is gravitational force.

  $F_e$ is electrical force.

  $r$ is distance between objects.

  $m_e$ is mass of electron.

  $m_p$ is mass of photon.

  $e$ is electrical charge.

Calculation:

The ratio of forces: $\frac{F_g}{F_e}=\frac{\left(\frac{G{m}_e{m}_p}{r^2}\right)}{\left(\frac{k{e}^2}{r^2}\right)}$

  $\therefore \frac{F_g}{F_e}=\frac{G{m}_e{m}_p}{k{e}^2}$

  $\therefore \frac{F_g}{F_e}=\frac{\left(\text{6}\text{.67}\times {\text{10}}^{-11}N\cdot m^2/k{g}^2\right)\left(9.11\times {10}^{-31}kg\right)\left(1.67\times {10}^{-27}kg\right)}{\left(9.00\times {\text{10}}^{9}N\cdot m^2/C^2\right){\left(1.60\times {10}^{-19}C\right)}^2}$

  $\therefore \frac{F_g}{F_e}=4.4\times {10}^{-40}$

Conclusion:

The ratio of the gravitational forces and electrical forces of a electron in hydrogen atom is $4.4\times {10}^{-40}$ . Yes, the gravitational force may be safely ignored.