Question

$\mathrm{N}=1.32\cdot {10}^6$Consider a hydrogen atom in stationary state n. (a) Show that the orbital period of an electron in quantum state n is $\mathrm{T}={\mathrm{n}}^3{\mathrm{T}}_1$and find a numerical value for${\mathrm{T}}_1.$ (b) On average, an atom stays in the n = 2 state for 1.6 ns before undergoing a quantum jump to the n = 1 state. On average, how many revolutions does the electron make before the quantum jump?

Short answer

(a) The numerical value is ${\mathrm{T}}_1=1.52\cdot {10}^{-16}\mathrm{s}$

(b) Evolutions does the electron go through before making the quantum leap is

Step-by-step solution

Part(a) Step 1: Given information

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

Part (a) Step 2: Finding orbital period of an electron

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

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

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

$$\begin{gathered} {\mathrm{v}}_{\mathrm{n}}=\frac{\mathrm{nħ}}{{\mathrm{mr}}_1} \\ {\mathrm{v}}_1=\frac{1\cdot 1.05\cdot {10}^{-34}}{9.11\cdot {10}^{-31}\cdot 0.0529\cdot {10}^{-9}}\text{(substitute)} \\ {\mathrm{v}}_1=2.19\cdot {10}^6\frac{\mathrm{m}}{\mathrm{s}} \\ {\mathrm{T}}_1=\frac{2{\mathrm{r}}_1\mathrm{\pi }}{{\mathrm{v}}_1} \\ {\mathrm{T}}_1=\frac{2\cdot 0.0529\cdot {10}^{-9}\cdot \mathrm{\pi }}{2.19\cdot {10}^6}\text{(substitute)} \\ {\mathrm{T}}_1=\boxed{1.52\cdot {10}^{-16}\mathrm{s}} \end{gathered}$$

Part (b) Step 2:

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

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

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" $\begin{array}{r}\mathrm{N}=\frac{{\mathrm{t}}_2}{{\mathrm{T}}_2}\\ \mathrm{N}=\frac{1.6\cdot {10}^{-9}}{1.216\cdot {10}^{-15}}\text{(substitute)}\\ \mathrm{N}=\boxed{1.32\cdot {10}^6}\end{array}$

Step 3: 

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