Question

Using the Bohr 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

Answer
We use the Rydberg formula to get the wavelength, and then we convert this to a frequency of $3.08\times {10}^{15}{s}^{-1}$
Explanation:
The Rydberg formula is
$\frac{1}{\lambda }=R_H(\frac{1}{n_i^2}-\frac{1}{n_i^2})$
where
$\lambda$ is the wavelength of the radiation,
$R_H$ is the Rydberg constant $(1.097\times {10}^{7}m6-1)$
$n_i$ and $n_f$are the initial and final energy levels
Thus,
$\frac{1}{\lambda }=1.097\times {10}^7{m}^{-1}(\frac{1}{4^2}-\frac{1}{1^2})=-1.028\times {10}^7{m}^{-1}$
$\lambda =-\frac{1}{1.028\times {10}^7{m}_1}=-9.723\times {10}^{-8}m$
The negative sign shows that energy is emitted.
Frequency and wavelength are related by the formula
$\lambda =c$
where
ν is the frequency of the lightc is the frequency of the light $(2.998\times {10}^{8}m.s^{-1})$
$\therefore v=\frac{c}{\lambda }=\frac{2.998\times {10}^{8}m.s^{-1}}{9.723\times {10}^{-8}m}=3.08\times {10}^{15}{s}^{-1}$