Chapter 16, Problem 59GP

To determine

To find: The radius of the electron’s orbit in hydrogen atom.

Answer to Problem 59GP

  $2.09\times {10}^{-10}\text{m}$

Explanation of Solution

Given:

Mass of the proton ${\mathrm{m}}_{\mathrm{p}}=1.67\times {10}^{-2}\text{kg}$

Mass of an electron ${\mathrm{m}}_{\mathrm{e}}=9.11\times {10}^{-31}\text{kg}$

Charge on a proton $\mathrm{p}=1.6\times {10}^{-19}\text{C}$

Charge on an electron $\mathrm{e}=-1.6\times {10}^{-19}\text{C}$

Orbital speed $\mathrm{v}=1.1\times {10}^6\text{m}/$ s

Speed of revolution of electron around proton in hydrogen atom $=1.1\times {10}^6\mathrm{m}/\mathrm{s}$

Orbit of revolution of electron is circular.

Formula used:

  $\mathrm{r}=\frac{\mathrm{k}{\mathrm{e}}^2}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}$

Calculation:

The radius of the orbit = r

Then the centripetal force acting on the electron = Electrostatic force of attraction,

Hence,

  $\mathrm{r}=\frac{\mathrm{k}{\mathrm{e}}^2}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}$

  $\begin{array}{l}\mathrm{r}=\frac{\mathrm{k}{\mathrm{e}}^2}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2}\\ \mathrm{r}=\frac{\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2}{\left(9.11\times {10}^{-31}\right){\left(1.1\times {10}^6\right)}^2}\text{m}\\ \mathrm{r}=2.09\times {10}^{-10}\text{m}\end{array}$

Conclusion:

Radius of electron’s orbit in hydrogen atom is given by $2.09\times {10}^{-10}\text{m}$ .