An electron initially in the n = 2 energy level in a hydrogen atom absorbs a photon of light with a frequency of 6.167 x 1014 s-1. Calculate the new energy level the electron will occupy.
I am not sure how to being.
the change in energy level is
... energy = Planck's Const. * freq
the Balmer Series starts at n=2
look at the energy levels to see how far the electron jumped
To solve this problem, we need to use the relationship between the energy of a photon of light and its frequency.
The energy of a photon can be calculated using the equation:
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.
In this problem, we are given the frequency of the absorbed photon as 6.167 x 10^14 s^-1. We can use this frequency value to calculate the energy of the absorbed photon.
E = (6.626 x 10^-34 J·s) * (6.167 x 10^14 s^-1)
Now, we know that when an electron absorbs a photon, it moves to a higher energy level. In the case of a hydrogen atom, the energy levels are quantized and are given by the equation:
E_n = -13.6eV/n^2
where E_n is the energy of the nth energy level, and n is the principal quantum number.
To find the new energy level, we need to find the value of n for which the energy of the absorbed photon matches the energy difference between the current energy level (n=2) and the new energy level.
Let's calculate the energy of the initial energy level (n=2):
E_2 = -13.6eV / (2^2)
Next, we subtract the energy of the initial energy level from the energy of the absorbed photon to find the energy difference:
ΔE = E - E_2
Finally, we can use the energy difference to find the new energy level by rearranging the formula for E_n:
n^2 = -13.6eV / ΔE
Solve for n to get the principal quantum number of the new energy level.