Question

a. What quantum number of the hydrogen atom comes closest to giving a 100-nm-diameter electron orbit?

b. What are the electron’s speed and energy in this state?

Short answer

(a) The radius of a hydrogen atom in its nth state$n=30.74\approx 31$

(b) In the nth state, the speed is${\mathrm{v}}_{31}=7.06\times {10}^4\mathrm{m}/\mathrm{s},{\mathrm{E}}_{31}=-0.44\mathrm{eV}$

Step-by-step solution

Part (a) Step 1 : Given information

The providing an electron orbit with a radius is 100 nano meters

Part (a) Step 2: Finding radius of a hydrogen atom 

We can begin by calculating the radius of a hydrogen atom in the nth state.

$\begin{array}{l}{\mathrm{r}}_{\mathrm{n}}={\mathrm{n}}^2{\mathrm{a}}_{\mathrm{B}}\\ {\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{D}}_{\mathrm{n}}}{2}\left(\text{radiusoverdiametar}\right)\\ {\mathrm{n}}^2=\frac{{\mathrm{r}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{B}}}\left(\text{express}{\mathrm{n}}^2\right)\\ \mathrm{n}=\sqrt{\frac{{\mathrm{D}}_{\mathrm{n}}}{2{\mathrm{a}}_{\mathrm{R}}}}\mathrm{n}=30.74\approx 31\left(\text{substitute}{\mathrm{r}}_{\mathrm{n}}\right)\\ \mathrm{n}=\sqrt{\frac{\left(100\times {10}^{-9}\right)}{2\cdot 0.0529\times {10}^{-9}}}\left(\mathrm{Substitute}\right)\\ \mathrm{n}=30.74\approx 31\end{array}$

Part (b) Step 2: Given information

The given Electronic speed and energy in Bohr Hydrogen atom

Part (b) Step 2: Finding Electric speed 

In the nth state, we may now utilize the expression:

$\begin{array}{l}{\mathrm{v}}_{\mathrm{n}}=\frac{{\mathrm{v}}_1}{\mathrm{n}}\left({\mathrm{v}}_1=2.19\times {10}^6\mathrm{m}/\mathrm{s}\right)\\ {\mathrm{v}}_{31}=\frac{{\mathrm{v}}_1}{31}\left(\text{substitute\hspace{0.17em}}\mathrm{n}=31\right)\\ {\mathrm{v}}_{31}=\frac{\left(2.19\times {10}^6\right)}{31}\left(\text{substitute\hspace{0.17em}}{\mathrm{v}}_1\right)\\ {\mathrm{v}}_{31}=\left(7.06\times {10}^4\right)\mathrm{m}/\mathrm{s}\\ {\mathrm{E}}_{\mathrm{n}}=-\frac{13.6\left(\mathrm{eV}\right)}{\mathrm{n}}\left(\text{energy\hspace{0.17em}in\hspace{0.17em}nth\hspace{0.17em}orbit}\right)\\ {\mathrm{E}}_{31}=-\frac{13.6\left(\mathrm{eV}\right)}{31}\left(\text{substitute\hspace{0.17em}}\mathrm{n}=31\right)\\ {\mathrm{E}}_{31}=-0.44\mathrm{eV}\end{array}$