A researcher observes hydrogen emitting photons of energy 10.2 eV.
What are the quantum numbers of the two states involved in the transition that emits these photons?
To determine the quantum numbers of the two states involved in the transition that emits these photons, you need to consider the energy difference between the initial and final states.
The energy of a photon can be calculated using the equation:
E = hv
Where E is the energy of the photon, h is Planck's constant (h ≈ 4.136 × 10^-15 eV·s), and v is the frequency of the photon.
Since we are given that the energy of the photon is 10.2 eV, we can convert it to joules by multiplying by the conversion factor: 1 eV ≈ 1.6 × 10^-19 Joules.
10.2 eV * (1.6 × 10^-19 J/eV) = 1.632 × 10^-18 J
Next, we can use the relationship between the energy of a photon and its frequency to find the frequency (v):
E = hv
h = 6.626 × 10^-34 J·s (Planck's constant, in Joules)
1.632 × 10^-18 J = (6.626 × 10^-34 J·s) * v
v = (1.632 × 10^-18 J) / (6.626 × 10^-34 J·s)
v ≈ 2.465 × 10^15 Hz
We know that the energy of a photon is related to the energy difference between the two states involved in the transition:
ΔE = E_final - E_initial
To determine the quantum numbers, we need to use the Balmer series equation for hydrogen atoms:
1/λ = R_H * (1/n_final^2 - 1/n_initial^2)
Where λ is the wavelength of the photon, R_H is the Rydberg constant for hydrogen (R_H ≈ 1.097 × 10^7 m^-1), and n_final and n_initial are the principal quantum numbers of the final and initial states, respectively.
The speed of light equation relates the frequency and wavelength:
c = λ * v
Where c is the speed of light (c ≈ 3 × 10^8 m/s).
Therefore, the wavelength (λ) of the photon can be calculated as follows:
λ = c / v
λ ≈ (3 × 10^8 m/s) / (2.465 × 10^15 Hz)
λ ≈ 1.215 × 10^-7 m
Now, we can use the wavelength to find the quantum numbers. Rearranging the Balmer series equation, we have:
1/λ = R_H * (1/n_final^2 - 1/n_initial^2)
Plugging in the values:
(1/1.215 × 10^-7 m) = (1.097 × 10^7 m^-1) * (1/n_final^2 - 1/n_initial^2)
Simplifying the equation and solving for the quantum numbers:
1/n_final^2 - 1/n_initial^2 = (1.215 × 10^-14)
To determine the possible quantum numbers, we can start by considering the Balmer series transitions.
For the Balmer series, the initial quantum number is always n = 2 (since it starts from the second energy level). Thus:
1/n_final^2 - 1/2^2 = (1.215 × 10^-14)
1/n_final^2 - 1/4 = (1.215 × 10^-14)
Multiplying both sides of the equation by 4 * n_final^2 will bring the equation into a simpler form:
4 - n_final^2 = (4 * n_final^2) * (1.215 × 10^-14)
4 - n_final^2 = 4.86 × 10^-14 * n_final^2
Now, we can solve the quadratic equation for n_final:
n_final^2 * (4.86 × 10^-14 + 1) - 4 = 0
Simplifying the equation further, we get:
n_final^2 * (5.86 × 10^-14) = 4
Dividing both sides by (5.86 × 10^-14), we find:
n_final^2 = 4 / (5.86 × 10^-14)
n_final^2 ≈ 6.83 × 10^13
Taking the square root of both sides:
n_final ≈ √(6.83 × 10^13)
n_final ≈ 8.27 × 10^6
Since n_final represents a principal quantum number and is expected to be a positive integer, we can conclude that n_final is approximately equal to 8.
Therefore, the two quantum numbers involved in the transition that emits these photons are n_initial = 2 (the initial state) and n_final = 8 (the final state).