Find the equation that describes the energy of a photon emitted as an electron relaxes from a higher energy level to a lower one in a hydrogen atom.
6 to 1
6 to 2
6 to 3
6 to 4
6 to 5
5 to 1
5 to 2
5 to 3
5 to 4
4 to 1
All of these? The first one is
E = 2.180E-18J x (1/1^2 - 1/6^2)
Is this the equation you use for each one, but just switch out the 6 and 1?
To find the equation that describes the energy of a photon emitted as an electron relaxes from a higher energy level to a lower one in a hydrogen atom, we need to use the equation for the energy of a photon. The equation is given by:
E = hf
Where:
E is the energy of the photon
h is Planck's constant (6.626 x 10^-34 J·s)
f is the frequency of the photon
The frequency of the photon can be calculated using the formula:
f = (c / λ)
Where:
c is the speed of light (3.00 x 10^8 m/s)
λ is the wavelength of the photon
In the case of hydrogen atoms, the energy levels can be represented by the formula:
E = -13.6eV / n^2
Where:
E is the energy of the electron in the hydrogen atom
n is the principal quantum number (corresponds to energy level)
Now, let's calculate the energies and frequencies of the photons for each transition and find the corresponding equations:
1. From energy level 6 to 1:
E6 = -13.6 eV / 6^2
E1 = -13.6 eV / 1^2
Calculate the energy difference: ΔE = E6 - E1
Calculate the frequency: f = ΔE / h
Finally, substitute the frequency into the equation E = hf to get the equation that describes the energy of the photon.
Repeat the above steps for all the given transitions (6 to 2, 6 to 3, 6 to 4, 6 to 5, 5 to 1, 5 to 2, 5 to 3, 5 to 4, 4 to 1) and you will obtain the equations that describe the energy of the photons emitted in each transition.