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)
I'm not sure how to do this; if someone could explain how to work it for one of the ni's, that would be a great help! Thank you! :)
410,434,486,657
To calculate the wavelengths of the visible lines in the line spectrum of hydrogen using the Rydberg equation, we can use the formula:
1/λ = R * (1/ni^2 - 1/nf^2)
where λ is the wavelength of the line, R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), ni is the initial energy level, and nf is the final energy level.
Let's calculate the wavelength for one of the initial energy levels, let's say ni = 4, with the final energy level nf = 2.
1/λ = R * (1/4^2 - 1/2^2)
1/λ = R * (1/16 - 1/4)
1/λ = R * (3/16)
Now, let's find the value of λ by taking the reciprocal on both sides:
λ = 16/3R
Substituting the value of R (1.097 x 10^7 m^-1) into the equation:
λ = 16/3 * (1.097 x 10^7 m^-1)
Calculating this expression gives us the wavelength in meters. However, we want the wavelength in nanometers, so we need to convert it:
λ = 16/3 * (1.097 x 10^7 m^-1) * (10^9 nm / 1 m)
Therefore, the wavelength for the transition from ni = 4 to nf = 2 is:
λ = (16/3) * (1.097 x 10^7) * (10^9) nm
You can repeat this calculation with different initial energy levels (ni values) to find the wavelengths for those transitions as well.
To calculate the wavelengths of the visible lines in the line spectrum of hydrogen using the Rydberg equation, we can use the formula:
1/λ = R * (1/nf^2 - 1/ni^2)
Where λ is the wavelength, R is the Rydberg constant, nf is the final energy level, and ni is the initial energy level.
In this case, nf is given as 2. We need to find the wavelengths for different initial energy levels (ni) of 3, 4, 5, and 6.
The Rydberg constant (R) for hydrogen is approximately 1.097 × 10^7 m⁻¹.
Let's calculate the wavelength for ni = 3 using the above formula.
1/λ = (1.097 × 10^7 m⁻¹) * (1/2^2 - 1/3^2)
Simplifying the equation:
1/λ = (1.097 × 10^7 m⁻¹) * (1/4 - 1/9)
Now, calculate:
1/λ = (1.097 × 10^7 m⁻¹) * (9/36 - 4/36)
= (1.097 × 10^7 m⁻¹) * (5/36)
= (5 × 1.097 × 10^7 m⁻¹) / 36
Finding λ:
λ = 36 / (5 × 1.097 × 10^7 m⁻¹)
= 36 / (5.485 × 10^7 m⁻¹)
Now, convert the wavelength to nanometers:
λ = (36 × 10^9 nm) / (5.485 × 10^7 m⁻¹)
≈ 656.29 nm
So, for ni = 3, the wavelength (λ) of the visible line in the line spectrum of hydrogen is approximately 656.29 nm.
Similarly, you can calculate the wavelengths for ni = 4, 5, and 6 using the same formula.
It just substitution into the equation.
1/wavelength = R(1/nf^2 - 1/ni^2)
1/wavelength = R(1/4 - 1/9) for ni-3
1/wavelength = R(1/4 - 1/16) for ni=4
etc.