Question

Using the Bohr’s model, how do you calculate the frequency of light in the hydrogen atom with the energy transition goes from $n = 4$ to $n = 1$.

Answer 1

Hint: The given problem can be solved by the formula that is used to determine the wavelength of the electron which is moving between the different levels of the atoms. The formula used here is also applicable for the spectrum of the different elements. By finding the value of wavelength the frequency can be found.

Formula used:
$ \dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right]$
Where,
$\lambda $ is the wavelength,
${R_H}$ is the Rydberg constant,
$n_f^{}$ Final state (lower state)
$n_i^{}$ Initial state (upper state)
$ c = \dfrac{1}{\lambda }$
Where,
$c$ is the frequency.
$\lambda $ is the Wavelength.

Complete step by step answer:
Bohr, in his atomic model retained the essential features of Rutherford’s model for hydrogen atom in which a single electron of mass $m$rotates around the nucleus of charge $ + e$. Bohr extended his theory and hypothesis by incorporating Planck’s quantum ideas.
The given problem can be solved with the help of the Rydberg formula. Rydberg formula that is used to determine the wavelength of the electron which is moving between the different levels of the atoms. The formula used here is also applicable for the spectrum of the different elements.

The formula can be mathematically represented as,
$ \dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right]$
Where,
$\lambda $ is the wavelength,
${R_H}$ is the Rydberg constant,
$n_f^{}$ Final levels (lower state),
$n_i^{}$ Initial levels (upper state).
To find the frequency, the value for wavelength must be found and the wavelength can be found using the Rydberg formula.
The values are given. The value of final level is $n = 4$, the value of initial level is $n = 1$, the Rydberg constant is $1.097 \times {10^7}{m^{ - 1}}$.
The formula we have.
$ \Rightarrow \dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{n_f^2}} - \dfrac{1}{{n_i^2}}} \right]$

Let us substitute the given values in the formula,
\[ \Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}{m^{ - 1}}\left[ {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{1^2}}}} \right]\]
Let’s solve the bracket terms first using division.
\[ \Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}{m^{ - 1}}\left[ {\dfrac{1}{{16}} - \dfrac{1}{{{1^{}}}}} \right]\]
Simplify the bracket terms using the division we get,
\[ \Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}{m^{ - 1}}\left[ {0.0625 - 1} \right]\]
On further simplifications inside the bracket terms we ger,
\[ \Rightarrow \dfrac{1}{\lambda } = 1.097 \times {10^7}{m^{ - 1}} - \left( { - 0.9375} \right)\]
The given equation can be simplified using the subtraction we get,
\[ \Rightarrow \dfrac{1}{\lambda } = 2.0345 \times {10^7}{m^{ - 1}}\]
\[ \Rightarrow \lambda = \dfrac{1}{{2.0345 \times {{10}^7}{m^{ - 1}}}}\]
$ \Rightarrow \lambda = 0.491521 \times {10^7}$

The value of wavelength is found. Next we need to find the value of frequency. To find the frequency the formula is given as,
$ c = \dfrac{1}{\lambda }$
Where,
$c$ is the Frequency
$\lambda $ is the Wavelength.
Let us substitute the value of the wavelength in the formula.
$ \Rightarrow c = \dfrac{1}{{0.491251528 \times {{10}^7}{m^{ - 1}}}}$
Using the division to simplify we get,
$ \Rightarrow c = 2.03561{s^{ - 1}}$
Therefore, the value of the wavelength is $0.491521 \times {10^7}$ and the frequency is $2.03561{s^{ - 1}}$.

Note: According to the Bohr, the electron will always revolve around the nucleus in a certain definite circular orbit without radiating any energy. The possible orbits are called stationary states. The allowed states are those for which the orbital angular momentum of the electron is equal to the integral multiple of $\hbar $.