Question

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${\mathrm{T}}_1=1.52\times {10}^{-16}\mathrm{s}$

(b) The number of rotations an electron does before making a quantum leap is$\mathrm{N}=1.32\times {10}^6$

Step-by-step solution

Part (a) Step 1: Given information

An electron's orbital period in quantum state n is${\mathrm{T}}_1=1.52\times {10}^{-16}\mathrm{s}$

Part (b) Step 2:  Finding the orbital period of electron 

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

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

We know ${\mathrm{r}}_1$is 0.0529 nm, thus we only need to find ${\mathrm{v}}_1$with the following expression:

localid="1651154704956" $$\begin{gathered} {\mathrm{v}}_{\mathrm{n}}=\frac{{\mathrm{mr}}_1}{\mathrm{m}} \\ {\mathrm{v}}_1=\frac{1\times \left(1.05\times {10}^{-34}\right)}{\left(9.11\times {10}^{-31}\right)\times \left(0.0529\times {10}^{-9}\right)}\left(\mathrm{substitute}\right) \\ {\mathrm{v}}_1=2.19\times {10}^6\frac{\mathrm{m}}{\mathrm{s}} \\ {\mathrm{T}}_1=\frac{2{\mathrm{r}}_1\mathrm{\pi }}{{\mathrm{v}}_1} \\ {\mathrm{T}}_1=\frac{\left(2\cdot 0.0529\times {10}^{-9}\right)\times \mathrm{\pi }}{\left(2.19\times {10}^6\right)}\left(\mathrm{substitute}\right) \\ {\mathrm{T}}_1=1.52\times {10}^{-16}\mathrm{s} \end{gathered}$$

Part (b) Step 1: Given information

An atom spends an average of 1.6 nanoseconds in the n = 2 state before quantum leaping to the n = 1 state.

Part (b) Step 2: Calculations

Part (a) orbital 's period expression can now be used:

$$\begin{gathered} {\mathrm{T}}_{\mathrm{n}}={\mathrm{n}}^3{\mathrm{T}}_1 \\ {\mathrm{T}}_2=2^3\times 1.52\times {10}^{-16}\left(\mathrm{Substitute}\right) \\ {\mathrm{T}}_2=1.216\times {10}^{-15}\mathrm{s} \end{gathered}$$

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

localid="1651154737208" $$\begin{gathered} \mathrm{N}=\frac{{\mathrm{t}}_2}{{\mathrm{T}}_2} \\ \mathrm{N}=\frac{\left(1.6\times {10}^{-9}\right)}{\left(1.216\times {10}^{-15}\right)}\left(\mathrm{substitute}\right) \\ \mathrm{N}=1.32\times {10}^6 \end{gathered}$$