Question

A hydrogen atom is in a state with energy $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{51}\mathbf{}\mathbf{eV}$. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

Short answer

The angular momentum of the electron in the atom with respect to an axis at the nucleus is $3.16\times {10}^{-34}\mathrm{kg}\cdot {\mathrm{m}}^2/\mathrm{s}$

Step-by-step solution

Define the angular momentum and energy.

If an electron with mass$\mathit{m}$moving with speed${\mathit{\upsilon }}_{\mathbf{n}}$in an orbit having the principal quantum number equal torole="math" localid="1664010570703" $\mathit{n}$and radius, then the minimum angular momentum${\mathit{L}}_{\mathbf{n}}$is:

$\begin{array}{l}\mathbf{L}\mathbf{=}\mathbf{m}{\mathbf{v}}_{\mathbf{n}}{\mathbf{r}}_{\mathbf{n}}\\ \mathbf{=}\mathbf{n}\left(\frac{h}{2\pi }\right)\end{array}$

Where, the value of Planck’s constant $\mathit{h}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{}\mathbf{J}\mathbf{\cdot }\mathbf{s}$.

In the Bohr model for the hydrogen atom the equations energy ${\mathit{E}}_{\mathbf{n}}$of the $\mathit{n}\mathit{t}\mathit{h}$orbit is:

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{e}\mathbf{V}}{{\mathbf{n}}^{\mathbf{2}}}$

Determine the angular momentum.

Given that, energy of hydrogen atom is$-1.51\mathrm{eV}$.

The principal quantum number$n$is:

$$\begin{gathered} n=\sqrt{\frac{-13.60\mathrm{eV}}{E_n}} \\ =\sqrt{\frac{-13.60}{-1.51}} \\ =3.00 \end{gathered}$$

Now, the angular momentum of electron is:

$$\begin{gathered} L=n\left[\frac{h}{2\pi }\right] \\ =3.00\left[\frac{6.626\times {10}^{-34}}{2\pi }\right] \\ =3.16\times {10}^{-34}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s} \end{gathered}$$

Hence, the angular momentum of the electron in the atom with respect to an axis at the nucleus is $3.16\times {10}^{-34}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}$.