The spectrum from a hydrogen vapour lamp is measured and four lines in the visible light range are observed. These lines are the so-called Balmer series, where an electron makes a transition from a higher level to the second energy level (nl=2).In this series, the transition from nu= 3 to nl= 2 produces the photon with the lowest energy, this corresponds to the line with the longest wavelength. This is measured to be λ = 646.9 nm.What value of the Rydberg constant R is obtained using these measurements?Express you answer in µm-1 to two decimal places
Question
The spectrum from a hydrogen vapour lamp is measured and four lines in the visible light range are observed. These lines are the so-called Balmer series, where an electron makes a transition from a higher level to the second energy level (nl=2).In this series, the transition from nu= 3 to nl= 2 produces the photon with the lowest energy, this corresponds to the line with the longest wavelength. This is measured to be λ = 646.9 nm.What value of the Rydberg constant R is obtained using these measurements?Express you answer in µm-1 to two decimal places
Solution
To find the Rydberg constant (R), we can use the formula for the wavelength of light emitted during electron transitions in the hydrogen atom:
1/λ = R * (1/nl^2 - 1/nu^2)
where λ is the wavelength, nl is the lower energy level, nu is the upper energy level, and R is the Rydberg constant.
Given that λ = 646.9 nm = 0.6469 µm, nl = 2, and nu = 3, we can substitute these values into the formula:
1/0.6469 = R * (1/2^2 - 1/3^2)
Solving for R, we get:
R = 1/0.6469 / (1/2^2 - 1/3^2)
R = 1.545 µm^-1
So, the Rydberg constant R obtained using these measurements is 1.55 µm^-1 to two decimal places.
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