Question

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the $\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}$and $\mathbf{3}$levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}\mathbf{}}\mathbf{s}\mathbf{.}$In the Bohr model, how many orbits does an electron in the $\mathbf{n}\mathbf{=}\mathbf{2}$level complete before returning to the ground level?

Short answer

(a) The speed of the electron in a hydrogen atom in the $\mathrm{n}=1,2\mathrm{and}3$levels are

role="math" localid="1664025558506" $2.18\times {10}^6\mathrm{m}/\mathrm{s},1.09\times {10}^6\mathrm{m}/\mathrm{s}\mathrm{and}7.27\times {10}^5\mathrm{m}/\mathrm{s}$respectively.

(b) The orbital period in $\mathrm{n}=1,2\mathrm{and}3\mathrm{levels}\mathrm{are}1.53\times {10}^{-16}\mathrm{s},1.22\times {10}^{-15}\mathrm{s},\mathrm{and}4.13\times {10}^{-15}\mathrm{s}.$.

(c) The number of orbitals of an electron in the $\mathrm{n}=2$level before returning to the ground level is $8.18\times {10}^6$.

Step-by-step solution

Define the Bohr model equations for the hydrogen atom.

In the Bohr model for the hydrogen atom the equations of radius ${\mathbf{r}}_{\mathbf{n}}$, time ${\mathbf{T}}_{\mathbf{n}}$and speed ${\mathbf{U}}_{\mathbf{n}}$of the $\mathbf{nth}$orbit is:

$$\begin{gathered} {\mathbf{r}}_{\mathbf{n}}\mathbf{=}{\mathbf{\in }}_{\mathbf{0}}\frac{\mathbf{}{\mathbf{n}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}\mathbf{}}{{\mathbf{\pi me}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ {\mathbf{\upsilon }}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\in }}_{\mathbf{0}}}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{2}\mathbf{nh}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ {\mathbf{T}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{4}{\mathbf{\in }}_{\mathbf{0}}{\mathbf{n}}^{\mathbf{3}}{\mathbf{h}}^{\mathbf{3}}}{{\mathbf{me}}^{\mathbf{4}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

Determine the speeds.

For $\mathrm{n}=1$the speed of electron is:

$$\begin{gathered} u_1=\frac{1}{\left(8.854\times {10}^{-12}\right)}\frac{\left(1.60\times {10}^{-19}\right)\left(1^2\right)\left(6.626\times {10}^{-34}\right)}{2\left(1\right)\left(6.626\times {10}^{-34}\right)} \\ =2.18\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

For $n=2$the speed of electron is:

localid="1664029272330" $$\begin{gathered} u_2=\frac{1}{\left(8.854\times {10}^{-12}\right)}\frac{\left(1.60\times {10}^{-19}\right)\left(1^2\right){\left(6.626\times {10}^{-34}\right)}^2}{2\left(2\right)\left(6.626\times {10}^{-34}\right)} \\ =1.09\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

For $n=3$the speed of electron is:

localid="1664029335707" $$\begin{gathered} u_3=\frac{1}{\left(8.854\times {10}^{-12}\right)}\frac{\left(1.60\times {10}^{-19}\right)\left(1^2\right){\left(6.626\times {10}^{-34}\right)}^2}{2\left(3\right)\left(6.626\times {10}^{-34}\right)} \\ =7.27\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the electron in a hydrogen atom in the $n=1,2$and $3$levels are $2.18\times {10}^6\mathrm{m}/\mathrm{s}$ and $7.27\times {10}^5\mathrm{m}/\mathrm{s}$respectively.

Determine the orbital periods.

For $n=1$the orbital period of electron is:

role="math" localid="1664030305197" $$\begin{gathered} T_1=\frac{4{\left(8.854\times {10}^{-12}\right)}^2{\left(1\right)}^3{\left(6.626\times {10}^{-34}\right)}^3}{\left(9.11\times {10}^{-31}\right){\left(1.60\times {10}^{-19}\right)}^4} \\ =1.53\times {10}^{-16}s \end{gathered}$$

For $n=2$the orbital period of electron is:

$$\begin{gathered} T_2=\frac{4{\left(8.854\times {10}^{-12}\right)}^2{\left(2\right)}^3{\left(6.626\times {10}^{-34}\right)}^3}{\left(9.11\times {10}^{-31}\right){\left(1.60\times {10}^{-19}\right)}^4} \\ =1.22\times {10}^{-15}s \end{gathered}$$

For $n=3$the orbital period of electron is:

$$\begin{gathered} T_3=\frac{4{\left(8.854\times {10}^{-12}\right)}^2{\left(3\right)}^3{\left(6.626\times {10}^{-34}\right)}^3}{\left(9.11\times {10}^{-31}\right){\left(1.60\times {10}^{-19}\right)}^4} \\ =4.13\times {10}^{-16}s \end{gathered}$$

Hence, the orbital period in$n=1,2$and$3$levels are$1.53\times {10}^{-16}\mathrm{s},1.22\times {10}^{-15}\mathrm{s},$and$4.13\times {10}^{-15}s$

Determine the number of orbitals.

Given that average lifetime$t$of the first excited level of the hydrogen atom is$1.00\times {10}^{-8}\mathrm{s}$.

The average life of $n=2$level of a hydrogen atom is:

role="math" localid="1664030735502" $\begin{array}{l}N=\frac{t}{T_3}\\ =\frac{1.00\times {10}^{-8}}{1.22\times {10}^{-15}}\\ =8.18\times {10}^6\mathrm{orbit}\end{array}$

Hence, the number of orbitals of an electron in the $n=2$level before returning to the ground level is $8.18\times {10}^6$.