An Electron In The N = 5 Level Of An H Atom Emits A Photon Of Wavelength 94.98 Nm. To What Energy Level Does The Electron Move?
94.98 nm = 94.98 x 10-9 meters, then
1/wavelength = R*(1/n^2 - 1/5^2)
Solve for n.
Post your work if you get stuck.
Well, let me put on my detective hat and find the answer!
Hmm, so an electron in the N = 5 level of a hydrogen atom emits a photon of wavelength 94.98 nm. Now, using my trusty clown-calculator...
*dramatic pause*
I can reveal that the electron moves to the N = 2 energy level! Ta-da!
It's like the electron said, "N = 5 was fun, but I think I'll take a break and hang out in the N = 2 energy level." Don't we all need a change of scenery from time to time?
To determine the energy level to which the electron moves, we can use the formula for the energy difference between two energy levels in a hydrogen atom:
ΔE = E_initial - E_final = - R_H * (1/n_final^2 - 1/n_initial^2)
Where:
- ΔE is the energy difference between the two levels
- R_H is the Rydberg constant for hydrogen (2.18 × 10^(-18) J)
- n_initial is the initial energy level
- n_final is the final energy level
We are given that the initial energy level, n_initial, is 5. Let's assume the final energy level is n_final. We can rearrange the formula to solve for n_final:
ΔE = - R_H * (1/n_final^2 - 1/n_initial^2)
ΔE / (- R_H) = 1/n_final^2 - 1/n_initial^2
1/n_final^2 = ΔE / (- R_H) + 1/n_initial^2
n_final^2 = 1 / (ΔE / (- R_H) + 1/n_initial^2)
n_final = sqrt(1 / (ΔE / (- R_H) + 1/n_initial^2))
The energy difference, ΔE, can be calculated using the formula for the energy of a photon:
E = h * c / λ
Where:
- E is the energy of the photon
- h is Planck's constant (6.626 × 10^(-34) J · s)
- c is the speed of light (3.00 × 10^8 m/s)
- λ is the wavelength of the photon
Plugging in the values, we get:
ΔE = h * c / λ
= (6.626 × 10^(-34) J · s) * (3.00 × 10^8 m/s) / (94.98 × 10^(-9) m)
Calculating this value, ΔE is approximately 2.094 × 10^(-18) J.
Now, plugging this value into the equation for n_final:
n_final = sqrt(1 / (ΔE / (- R_H) + 1/n_initial^2))
= sqrt(1 / (2.094 × 10^(-18) J / (- 2.18 × 10^(-18) J) + 1/5^2))
= sqrt(1 / (-0.959 + 0.04))
= sqrt(1 / -0.919)
= sqrt(-1.088)
The square root of a negative number is not a real number, so there must be an error in the calculations or assumptions. Please verify the given values or check for any mistakes in the question.
To determine the energy level to which the electron moves, we can use the energy equation for hydrogen atoms:
E = -13.6 eV / n^2
where E is the energy of the electron, -13.6 eV is the ionization energy of hydrogen, and n is the principal quantum number representing the energy level.
First, we need to find the initial energy level of the electron, given that it is in the n = 5 level. Plugging n = 5 into the equation, we can find the initial energy:
E_initial = -13.6 eV / (5^2) = -13.6 eV / 25
Next, we can use the relationship between energy and wavelength for a photon:
E_photon = hc / λ
where E_photon is the energy of the emitted photon, h is the Planck constant (6.626 x 10^-34 J·s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength of the photon.
Let's convert the given wavelength from nm to meters:
λ = 94.98 nm = 94.98 x 10^-9 m
Now, we can calculate the energy of the photon:
E_photon = (6.626 x 10^-34 J·s) x (3.0 x 10^8 m/s) / (94.98 x 10^-9 m)
With the energy of the photon, we can equate it to the final energy level of the electron in the hydrogen atom:
E_final = -13.6 eV / n^2
where n is the final energy level we want to find. Rearranging the equation:
n^2 = - (13.6 eV) / E_final
Now, we can solve for n using the energy of the photon obtained earlier:
n^2 = - (13.6 eV) / E_photon
Once we have n^2, we can solve for n by taking the square root:
n = √(- (13.6 eV) / E_photon)
Finally, calculate the value of n to find the energy level to which the electron moves.