What is the frequency of the photons emitted by hydrogen atoms when they undergo transition from n=5 to n=3?
DrBob222 thank you so much dude
spent 57 hours working that out for my PhD, I now have it and am earning big robux. thank you dude. I hope you ahve a good life
To find the frequency of the photons emitted by hydrogen atoms during a transition from the n=5 to n=3 energy level, we can use the Rydberg formula:
1/λ = R_H * (1/n1^2 - 1/n2^2)
Where:
- λ is the wavelength of the emitted photon
- R_H is the Rydberg constant for hydrogen (approximately 1.097 × 10^7 m⁻¹)
- n1 and n2 are the initial and final energy levels, respectively.
To find the frequency, we need to convert the wavelength to frequency using the formula:
c = λ * ν
Where:
- c is the speed of light (approximately 3.00 × 10^8 m/s)
- ν is the frequency of the emitted photon.
Let's substitute the values into the equations to find the frequency step-by-step:
Step 1: Calculate the wavelength (λ)
1/λ = R_H * (1/n1^2 - 1/n2^2)
1/λ = 1.097 × 10^7 m⁻¹ * (1/5^2 - 1/3^2)
1/λ = 1.097 × 10^7 m⁻¹ * (1/25 - 1/9)
1/λ = 1.097 × 10^7 m⁻¹ * (9/225 - 25/225)
1/λ = 1.097 × 10^7 m⁻¹ * (-16/225)
λ = - 225 / (1.097 × 10^7 m⁻¹ * 16)
λ ≈ - 1.303 × 10^-8 m (negative sign indicates emission)
Step 2: Calculate the frequency (ν)
c = λ * ν
ν = c / λ
ν = (3.00 × 10^8 m/s) / (-1.303 × 10^-8 m)
ν ≈ - 2.301 × 10^16 Hz
Keep in mind that the negative sign indicates emission. So, the frequency of the photons emitted by hydrogen atoms during this transition is approximately 2.301 × 10^16 Hz.
To determine the frequency of the photons emitted by hydrogen atoms during this transition, we can use the formula for the energy of a photon:
E = hc/λ
Where:
- E is the energy of the photon
- h is Planck's constant (approximately 6.626 x 10^-34 joule-seconds)
- c is the speed of light in a vacuum (approximately 3 x 10^8 meters/second)
- λ is the wavelength of the photon.
The energy difference between two energy levels in a hydrogen atom is given by the formula:
ΔE = E2 - E1 = -13.6eV * (1/n1^2 - 1/n2^2)
Where:
- ΔE is the energy difference between the two energy levels
- E2 and E1 are the energies of the final and initial energy levels, respectively
- n2 and n1 are the principal quantum numbers of the final and initial energy levels, respectively
- -13.6 eV is the ionization energy of hydrogen.
For the transition from n=5 to n=3, we can substitute these values into the formula:
ΔE = -13.6eV * (1/3^2 - 1/5^2)
Simplifying the equation gives:
ΔE = -13.6eV * (1/9 - 1/25)
Next, we convert electron volts (eV) to joules (J) by using the conversion factor:
1eV = 1.6 x 10^-19 J
Substituting the values and converting:
ΔE = -13.6 * 1.6 x 10^-19 J * (1/9 - 1/25)
Calculating the energy difference gives:
ΔE = 2.04 x 10^-19 J
Now, we can substitute this ΔE value into the formula for the energy of the photon:
E = ΔE = hc/λ
Rearranging the equation to solve for the wavelength (λ) gives:
λ = hc/ΔE
Substituting the values:
λ = (6.626 x 10^-34 J*s * 3 x 10^8 m/s) / (2.04 x 10^-19 J)
The wavelength λ can be found by dividing the product of Planck's constant and the speed of light by the energy difference. By calculating this expression, we can determine the wavelength, which can then be used to find the frequency of the photon:
λ = (6.626 x 10^-34 J*s * 3 x 10^8 m/s) / (2.04 x 10^-19 J) = 9.71 x 10^-8 m
Finally, we can find the frequency (ν) of the photon using the equation:
ν = c/λ
Substituting the values:
ν = (3 x 10^8 m/s) / (9.71 x 10^-8 m)
By dividing the speed of light by the wavelength, we can determine the frequency:
ν ≈ 3.09 x 10^15 Hz
Therefore, the frequency of the photons emitted by hydrogen atoms when they undergo a transition from n=5 to n=3 is approximately 3.09 x 10^15 Hz (or 3.09 THz).
w = wavelength
1/w = 1.097E7(1/3^2 - 1/5^2)
convert w to f with c = wf