Calculate the allowed wavelengths for the emission of a photon initially in the n=3 state in a hydrogen atom
1/lambda= Rh (1/n1-1/n2)
and
1/lambda=Rh(1/n1 -1/n3)
and
1/lambda=Rh(1/n2-1/n3)
Rh=1.097E7
i understand i would use Rydberg formula but what is the values of n
To calculate the allowed wavelengths for the emission of a photon from a hydrogen atom in the n=3 state, we can use the equation for the energy of the electron in a hydrogen atom:
E = -13.6 eV / n²
In this equation, E represents the energy of the electron, -13.6 eV is the ionization energy of hydrogen, and n is the principal quantum number.
Since we are interested in the emission of a photon, we need to find the energy difference between two energy levels. For a photon to be emitted, the electron must transition from a higher energy level to a lower energy level.
In this case, the initial energy level is n=3. We can calculate the energy of the electron in this state using the formula:
E₁ = -13.6 eV / (3)²
E₁ = -13.6 eV / 9
E₁ = -1.511 eV
To find the allowed wavelengths, we need to consider the final energy levels the electron can transition to from the n=3 state. According to the Bohr model, the electron can transition to energy levels with a lower principal quantum number (n).
For transitions from n=3 to a lower energy level, we can calculate the energy difference (ΔE) between the final and initial states. This energy difference corresponds to the energy of the emitted photon:
ΔE = E₂ - E₁
Now, we can calculate the final energy (E₂) from the energy equation for hydrogen:
E = -13.6 eV / n²
For n=1, the final energy level is:
E₂ = -13.6 eV / (1)²
E₂ = -13.6 eV
The energy difference (ΔE) between the final and initial states is:
ΔE = -13.6 eV - (-1.511 eV)
ΔE = -13.6 eV + 1.511 eV
ΔE ≈ -12.089 eV
Now, we need to convert the energy difference to a wavelength. We can use the following relation:
ΔE = hc / λ
Where:
ΔE is the energy difference in Joules,
h is the Planck's constant (6.626 x 10^-34 J·s),
c is the speed of light (2.998 x 10^8 m/s),
and λ is the wavelength of the emitted photon.
Rearranging the equation, we can solve for λ:
λ = hc / ΔE
Plugging in the values:
λ = (6.626 x 10^-34 J·s) * (2.998 x 10^8 m/s) / (-12.089 eV)
Calculating this gives us the wavelength in meters.