Calulate the wavelengths (in nm) of the visible lines in the line spectrum of hydrogen using the Rydberg equation (nf = 2; ni = 3, 4, 5, and 6).
Isn't this just a straight substitution into the Rydberg equation?
410 nm
I am student
To calculate the wavelengths of the visible lines in the line spectrum of hydrogen using the Rydberg equation, you need to use the formula:
1/λ = R * (1/nf^2 - 1/ni^2)
where λ represents the wavelength, R is the Rydberg constant, nf is the final energy level, and ni is the initial energy level.
The Rydberg constant for hydrogen is given as R = 1.097373 * 10^7 m^-1.
Now, let's calculate the wavelengths for each initial energy level (ni) provided (ni = 3, 4, 5, and 6) and nf = 2.
For ni = 3:
1/λ = R * (1/2^2 - 1/3^2)
1/λ = R * (1/4 - 1/9)
1/λ = R * (9/36 - 4/36)
1/λ = R * (5/36)
λ = 36/5 * 1/R
λ ≈ 36/5 * 1 / (1.097373 * 10^7)
λ ≈ 6.565 nm
For ni = 4:
1/λ = R * (1/2^2 - 1/4^2)
1/λ = R * (1/4 - 1/16)
1/λ = R * (4/16 - 1/16)
1/λ = R * (3/16)
λ = 16/3 * 1/R
λ ≈ 16/3 * 1 / (1.097373 * 10^7)
λ ≈ 4.897 nm
For ni = 5:
1/λ = R * (1/2^2 - 1/5^2)
1/λ = R * (1/4 - 1/25)
1/λ = R * (25/100 - 4/100)
1/λ = R * (21/100)
λ = 100/21 * 1/R
λ ≈ 100/21 * 1 / (1.097373 * 10^7)
λ ≈ 1.850 nm
For ni = 6:
1/λ = R * (1/2^2 - 1/6^2)
1/λ = R * (1/4 - 1/36)
1/λ = R * (9/36 - 1/36)
1/λ = R * (8/36)
λ = 36/8 * 1/R
λ ≈ 36/8 * 1 / (1.097373 * 10^7)
λ ≈ 4.133 nm
Therefore, the wavelengths (in nm) for the visible lines in the hydrogen spectrum are approximately:
- 6.565 nm
- 4.897 nm
- 1.850 nm
- 4.133 nm