91Ó°ÊÓ

The orbital period of an electron in quantum state n is$\mathrm{T}={\mathrm{n}}^3{\mathrm{T}}_1$

To begin, we can use the following expression for the orbital period of an electron in the n state:

$$\begin{gathered} {\mathrm{T}}_{\mathrm{n}}=\frac{2{\mathrm{r}}_{\mathrm{n}}\mathrm{Ï€}}{{\mathrm{v}}_{\mathrm{n}}} \\ {\mathrm{r}}_{\mathrm{n}}={\mathrm{r}}_1{\mathrm{n}}^2\text{(orbital radius)} \\ {\mathrm{v}}_{\mathrm{n}}=\frac{{\mathrm{v}}_1}{\mathrm{n}}\text{(orbital speed)} \\ {\mathrm{T}}_{\mathrm{n}}=\frac{2{\mathrm{r}}_1{\mathrm{n}}^2\mathrm{Ï€}}{\frac{{\mathrm{v}}_1}{\mathrm{n}}}\left(\text{substitute}{\mathrm{r}}_{\mathrm{n}}\text{and}{\mathrm{v}}_{\mathrm{n}}\right) \\ {\mathrm{T}}_{\mathrm{n}}=\frac{2{\mathrm{r}}_1{\mathrm{Ï€²Ô}}^3}{{\mathrm{v}}_1}=\boxed{{\mathrm{n}}^3{\mathrm{T}}_1}\left(\mathrm{edit}\right) \end{gathered}$$

we know that ${\mathrm{r}}_1\mathrm{is}0.0529\mathrm{nm},$we can only use${\mathrm{v}}_1$the following expression to find:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{n}}=\frac{\mathrm{²ÔÄ\S }}{{\mathrm{mr}}_1} \\ {\mathrm{v}}_1=\frac{1â‹\dots 1.05â‹\dots {10}^{−34}}{9.11â‹\dots {10}^{−31}â‹\dots 0.0529â‹\dots {10}^{−9}}\text{(substitute)} \\ {\mathrm{v}}_1=2.19â‹\dots {10}^6\frac{\mathrm{m}}{\mathrm{s}} \\ {\mathrm{T}}_1=\frac{2{\mathrm{r}}_1\mathrm{Ï€}}{{\mathrm{v}}_1} \\ {\mathrm{T}}_1=\frac{2â‹\dots 0.0529â‹\dots {10}^{−9}â‹\dots \mathrm{Ï€}}{2.19â‹\dots {10}^6}\text{(substitute)} \\ {\mathrm{T}}_1=\boxed{1.52â‹\dots {10}^{−16}\mathrm{s}} \end{gathered}$$

Part (b) Step 2:

Now we can utilize the orbital period expression from section a):

role="math" localid="1650806735056" $\begin{array}{r}{\mathrm{T}}_{\mathrm{n}}={\mathrm{n}}^3{\mathrm{T}}_1\\ {\mathrm{T}}_2=2^3â‹\dots 1.52â‹\dots {10}^{−16}\text{(substitute)}\\ {\mathrm{T}}_2=1.216â‹\dots {10}^{−15}\mathrm{s}\end{array}$

We can calculate the number of rotations an electron does before making a quantum jump by dividing the time spent in the n=2 state by the orbital period${\mathrm{T}}_2$.

role="math" localid="1650806935356" $\begin{array}{r}\mathrm{N}=\frac{{\mathrm{t}}_2}{{\mathrm{T}}_2}\\ \mathrm{N}=\frac{1.6â‹\dots {10}^{−9}}{1.216â‹\dots {10}^{−15}}\text{(substitute)}\\ \mathrm{N}=\boxed{1.32â‹\dots {10}^6}\end{array}$

$$\begin{gathered} \left(a\right)T_1=1.52â‹\dots {10}^{−16}\mathrm{s} \\ \left(\mathrm{b}\right)\mathrm{N}=1.32â‹\dots {10}^6 \end{gathered}$$