determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=2 and absorbs a photon of light with a frequency of 4.57 x 10^14 Hz.
E = h*frequency
Solve for E.
E = 2.180E-18*(1/4 - 1/x^2)
Solve for x which is the end orbit for the electron. The 1/4 comes from 1/n^2 - 1/2^2 = 1/4.
I looked at it quickly; I think x = 3.
Why did the electron bring a flashlight to the party? Because it wanted to find its "n" value! Alright, let's shed some light on this question!
In a hydrogen atom, the transition between energy levels is determined by the equation: ΔE = hf, where ΔE is the change in energy, h is the Planck's constant (approximately 6.626 x 10^-34 Joule-seconds), and f is the frequency of the absorbed photon.
To determine the end (final) value of n, we first need to find the change in energy. In the case of a hydrogen atom, the energy change is given by: ΔE = -13.6 eV * (1/n^2 final - 1/n^2 initial), where -13.6 eV is the energy of an electron in the first energy level (n=1).
Given that the electron starts in n=2, the initial value of n is 2. We can rearrange the energy equation to solve for the final value of n:
ΔE = -13.6 eV * (1/n^2 final - 1/2^2) = hf
Let's plug in the values. The energy change, ΔE, is given by -13.6 eV * (1/n^2 final - 1/2^2). The frequency, f, is 4.57 x 10^14 Hz.
Now, this equation might take a few steps to solve, so let me grab my thinking clown cap and calculate it for you. Give me a moment!
To determine the final value of n, we need to use the formula for the energy of a photon:
E = hf
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the photon.
First, we need to calculate the energy of the photon:
E = (6.626 x 10^-34 J·s) * (4.57 x 10^14 Hz)
E ≈ 3.019 x 10^-19 J
Next, we can use the formula for the energy of an electron in the hydrogen atom:
E = -13.6 eV * (1/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.
We rearrange this formula to solve for n:
n^2 = -13.6 eV / E
n^2 = -13.6 eV / (3.019 x 10^-19 J)
Now, let's convert the energy from joules to electron volts:
1 eV = 1.602 x 10^-19 J
E = (3.019 x 10^-19 J) / (1.602 x 10^-19 J/eV)
E ≈ 1.886 eV
Substituting this value back into the equation:
n^2 = -13.6 eV / 1.886 eV
n^2 ≈ 7.2
Taking the square root of both sides:
n ≈ √7.2
n ≈ 2.68
Since the principal quantum number must be a whole number, we round the result to the nearest whole number:
n ≈ 3
Therefore, the final value of n is 3.
To determine the final value of n in a hydrogen atom transition, we can use the equation that relates the frequency of the absorbed photon to the difference in energy levels of the hydrogen atom:
ΔE = hf
Where:
ΔE = Change in energy level
h = Planck's constant (6.626 x 10^-34 J·s)
f = Frequency of the absorbed photon
In this case, since the electron starts in n=2 and absorbs a photon with a frequency of 4.57 x 10^14 Hz, we can calculate the change in energy level:
ΔE = hf = (6.626 x 10^-34 J·s)(4.57 x 10^14 Hz)
To find the final value of n, we need to use the Rydberg formula, which relates the change in energy level to the initial and final energy levels of the electron:
ΔE = 13.6 eV * [1/n_final^2 - 1/n_initial^2]
Where:
ΔE = Change in energy level (in electron volts, eV)
n_initial = Initial energy level
n_final = Final energy level
Since we are looking for the final value of n, we can rearrange this equation to solve for n_final:
1/n_final^2 = 1/n_initial^2 + ΔE / 13.6 eV
Plugging in the values for ΔE and n_initial (n_initial = 2), we can solve for n_final.