What is the frequency of the photons emitted by hydrogen atoms when they undergo transition from n=4 to n=3?
1/wavelength = R(1/(n1)^2 - 1/(n2)^2
The first term is 3^2 and the second is 4^2. R you can find in your book or on the web.
To determine the frequency of the photons emitted by hydrogen atoms during the transition from one energy level to another, we can use the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \(\lambda\) is the wavelength of the emitted photon
- \(R\) is the Rydberg constant (\(R = 1.097 \times 10^7 \, \text{m}^{-1}\))
- \(n_1\) and \(n_2\) are the initial and final energy levels, respectively.
In this case, the hydrogen atom undergoes a transition from \(n_1 = 4\) to \(n_2 = 3\). Substituting these values into the formula:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1} \left( \frac{1}{4^2} - \frac{1}{3^2} \right) \]
Now we can calculate the value of \(\lambda\).
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1} \left( \frac{1}{16} - \frac{1}{9} \right) \]
To find the frequency (f) of the emitted photon, we can use the equation:
\[ f = \frac{c}{\lambda} \]
Where:
- \(c\) is the speed of light (\(c = 3.00 \times 10^8 \, \text{m/s}\))
Plugging in the calculated value of \(\lambda\):
\[ f = \frac{3.00 \times 10^8 \, \text{m/s}}{\lambda} \]
Now we can find the frequency (f) of the emitted photons.