Question

Calculate the frequency difference between the n = 4 → n = 2 transitions in regular hydrogen (one proton) and tritium (one proton and two neutrons). Unlike in problem 3, you can ignore the atom’s recoil. (This is called the isotope shift and is an important tool to learn about the size of nuclei.)

Answer 1

Answer 2

To calculate the frequency difference between the n = 4 → n = 2 transitions in regular hydrogen and tritium, we can use the Rydberg formula, which gives the frequency (or energy) of a photon emitted or absorbed during a transition between energy levels in a hydrogen-like atom:

1 / λ = R * (1/n_f^2 - 1/n_i^2)

where: λ is the wavelength of the emitted or absorbed photon, R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), n_f is the final energy level (higher principal quantum number), n_i is the initial energy level (lower principal quantum number).

For the n = 4 → n = 2 transition in regular hydrogen: n_f = 2 n_i = 4

1 / λ_hydrogen = R * (1/2^2 - 1/4^2) 1 / λ_hydrogen = R * (1/4 - 1/16) 1 / λ_hydrogen = R * (3/16)

Now, let's calculate the frequency difference for tritium, which has one proton and two neutrons:

For the n = 4 → n = 2 transition in tritium: n_f = 2 n_i = 4

1 / λ_tritium = R * (1/2^2 - 1/4^2) 1 / λ_tritium = R * (1/4 - 1/16) 1 / λ_tritium = R * (3/16)

The frequency difference (Δν) between the transitions can be calculated as follows:

Δν = ν_tritium - ν_hydrogen Δν = (c / λ_tritium) - (c / λ_hydrogen)

Where c is the speed of light (approximately 3.00 x 10^8 m/s).

Δν = (3.00 x 10^8 m/s) * (16/3R - 16R) Δν = 16/3R * (3.00 x 10^8 m/s - R * 16)

Now, we need to use the value of the Rydberg constant R:

R = 1.097 x 10^7 m^-1

Δν = 16/3 * (3.00 x 10^8 m/s - 1.097 x 10^7 m^-1 * 16) Δν = 16/3 * (3.00 x 10^8 m/s - 1.7552 x 10^8 m/s) Δν = 16/3 * (1.2448 x 10^8 m/s) Δν = 6.996 x 10^7 m/s

The frequency difference between the n = 4 → n = 2 transitions in regular hydrogen and tritium is approximately 6.996 x 10^7 m/s.