what is the wavelength of a photon emitted during a transition from the initial state(ni)=5 state to the final state(nf)=2 state in the hydrogen atom?
1/wavelength = R[1(2)^2 - (1/5)^2]
R is the Rydberg constant. You can find that on Google. It's about 1.097E7
Post your work if you get stuck.
To find the wavelength of a photon emitted during a transition from the initial state (ni) to the final state (nf) in the hydrogen atom, we can use the Rydberg formula.
The Rydberg formula for the calculation of the wavelength (λ) of a photon emitted during a transition in the hydrogen atom is given as follows:
1/λ = R * (1/nf^2 - 1/ni^2)
Where:
- λ is the wavelength of the photon
- R is the Rydberg constant (approximately 1.097 × 10^7 m^-1)
- nf is the principal quantum number of the final state
- ni is the principal quantum number of the initial state
In this case, the initial state (ni) is 5, and the final state (nf) is 2.
Plugging these values into the formula, we have:
1/λ = R * (1/2^2 - 1/5^2)
Simplifying further:
1/λ = R * (1/4 - 1/25)
1/λ = R * (25/100 - 4/100)
1/λ = R * (21/100)
Now, let's substitute the value of the Rydberg constant:
1/λ = (1.097 × 10^7 m^-1) * (21/100)
1/λ = 2.301 × 10^6 m^-1
Taking the reciprocal of both sides:
λ = 1 / (2.301 × 10^6 m^-1)
λ ≈ 4.347 × 10^-7 meters
Therefore, the wavelength of the photon emitted during the transition from the initial state (ni) = 5 to the final state (nf) = 2 in the hydrogen atom is approximately 4.347 × 10^-7 meters.