The lines in the visible portion of the hydrogen spectrum are called the Balmer series, for which m=2 in the Rydberg equation. Calculate the wavelength in nanometers of the spectral line in this series for which n=4.
delta E = 2.180E-18 J(1/a^2 - 1/b^2) where a = 2 and b= 4) so
dE = 2.180E-18 J (1/2^2 - 1/4^2)
Solve for dE. Then dE = hc/wavelength. h is Planck's constant, c is speed of light. Solve for wavelength (in meters) and convert to nanometers. Post your work if you get stuck.
So for dE I got 4.0875 × 10-19 joules
Over
(6.62607004 × 10-34 m2 kg / s)(299 792 458 m / s)
Do I have to then covert into nm?
To calculate the wavelength in nanometers for a spectral line in the Balmer series with n = 4, we can use the Rydberg equation. The Rydberg equation is given by:
1/λ = R * ((1/n₁²) - (1/n₂²))
Where:
- λ is the wavelength of the spectral line
- R is the Rydberg constant (1.097 × 10^7 m⁻¹)
- n₁ and n₂ are the initial and final energy levels of the electron
In this case, for the Balmer series, m=2. The Balmer series corresponds to the transitions where the electron starts from the higher energy level (n ≥ 3) and ends at the m = 2 level.
Substituting n₁ = 4 and n₂ = 2 into the Rydberg equation:
1/λ = R * ((1/4²) - (1/2²))
Simplifying the equation:
1/λ = R * (1/16 - 1/4)
1/λ = R * (1/16 - 4/16)
1/λ = R * (-3/16)
Rearranging the equation to solve for λ:
λ = 1 / (R * (-3/16))
Now we can calculate the value of λ:
λ = -16 / (3 * R)
λ = -16 / (3 * 1.097 × 10^7 m⁻¹)
Calculating this value gives us the wavelength in meters. To convert it to nanometers, we can multiply it by 10^9:
λ = (-16 × 10^9) / (3 * 1.097 × 10^7) nm
Evaluating this equation will give us the wavelength in nanometers for the spectral line in the Balmer series with n = 4.