Question

In the Bohr model of the hydrogen atom, the electron can have only certain velocities. Obtain a formula for the allowed velocities, then obtain a numerical value for the highest speed possible.

Short answer

The equation for the allowed velocity of the electron is $\frac{q^2}{4\pi {\epsilon }_0/n}$and the numeric value of the maximum allowed velocity is_{role="math" localid="1659758061348" $2.18\times {10}^6\mathrm{m}/\mathrm{s}$}.

Step-by-step solution

Bohr Model.

The expression for the velocity for the electron in the Bohr model of hydrogen is given by,

${\mathit{v}}^{\mathbf{2}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{m}\mathbf{r}}\frac{\mathbf{1}}{\mathbf{r}}$

The expression for the radius of the hydrogen atom of the Bohr model is given by,

$\mathit{r}\mathbf{=}\frac{\left(4{\mathrm{\pi \epsilon }}_0\right){\mathbf{d}}^{\mathbf{2}}{\mathbf{n}}^{\mathbf{2}}}{\mathbf{m}{\mathbf{q}}^{\mathbf{2}}}$

Use the expression of velocity and radius for calculation.

The expression for the velocity of the electron is evaluated as,

$$\begin{gathered} v^2=\frac{q^2}{4\pi {\epsilon }_{0}mr}\frac{1}{\left({\displaystyle \frac{\left(4\pi {\epsilon }_0\right)d^2{n}^2)}{m{q}^2}}\right)} \\ =\frac{q^2}{4\pi {\epsilon }_0/n} \end{gathered}$$

The velocity of the electron is calculated as,

role="math" localid="1659758442389" $$\begin{gathered} v=\frac{q^2}{4\pi {\epsilon }_{0}hn} \\ =\frac{{\left(1.6\times {10}^{-9}C\right)}^2}{4\mathrm{\pi }\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(1.0546\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(1\right)} \\ =2.18\times {10}^{6}m/s \end{gathered}$$

Therefore, the equation for the allowed velocity of the electron is$\frac{q^2}{4{\mathrm{\pi \epsilon }}_0/n}$and the numeric value of the maximum allowed velocity is_{$2.18\times {10}^{6}m/s$}.