What are the wavelengths, in nanometers, of the bright lines of the hydrogen emission spectrum corresponding to the transition: n=5 to n=2.
E = hc/wavelength = (1/4 - 1/25)
Thanks for that useless answer DrBob
To determine the wavelengths of the bright lines of the hydrogen emission spectrum corresponding to the transition from n=5 to n=2, you can use the Rydberg formula. The Rydberg formula describes the wavelengths of radiation emitted by hydrogen atoms in terms of the Rydberg constant (R), which is equal to 1.097 × 10^7 m^-1.
The formula is as follows:
1/λ = R * (1/n₁² - 1/n₂²)
Where:
- λ is the wavelength of light emitted
- R is the Rydberg constant
- n₁ and n₂ are the initial and final energy levels of the electron
In this case, the initial energy level of the electron is n₁ = 5 and the final energy level is n₂ = 2.
Plugging these values into the formula, we get:
1/λ = R * (1/5² - 1/2²)
Simplifying further:
1/λ = R * (1/25 - 1/4)
1/λ = R * (4/100 - 25/100)
1/λ = R * (-21/100)
1/λ = -R * (21/100)
Finally, solving for λ:
λ = 100 / (-R * 21)
Using the value of R = 1.097 × 10^7 m^-1, we can calculate:
λ = 100 / (-1.097 × 10^7 * 21) (in meters)
Converting the result into nanometers (1 meter = 10^9 nanometers), we get:
λ ≈ 434.168 nm
Therefore, the wavelength (λ) of the bright lines of the hydrogen emission spectrum corresponding to the n=5 to n=2 transition is approximately 434.168 nanometers.