An hydrogen atom in the ground state absorbs a photon with a wavelength one and after release a photon with a wavelength two,on which orbital will drop the electron?
To determine on which orbital the electron will drop after an hydrogen atom absorbs a photon with a wavelength of one and then releases a photon with a wavelength of two, we can use the concept of energy levels and transitions in the hydrogen atom.
The energy levels in a hydrogen atom are represented by the principal quantum number (n). The ground state is when the electron is in the lowest energy level, which corresponds to n = 1.
When an electron absorbs a photon, it gains energy and moves to a higher energy level. The energy difference between the initial and final energy levels is equal to the energy of the absorbed photon.
In this case, the hydrogen atom initially absorbs a photon with a wavelength of one (λ = 1). To determine the energy of the absorbed photon, we can use the equation:
E = hc/λ
Where:
E = energy of the photon
h = Planck's constant (approximately 6.626 x 10^-34 J·s)
c = speed of light (approximately 3.00 x 10^8 m/s)
Plugging in the values, we have:
E = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/(1)
After the absorbed photon increases the electron's energy, it will then drop back down to a lower energy level and release a photon with a wavelength of two (λ = 2). To determine the energy of the released photon, we can again use the equation:
E = hc/λ
Plugging in the values, we have:
E = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/(2)
Now, we need to compare the energy difference between the absorbed and released photons to determine the change in energy level for the electron.
If the energy difference is equal to the energy level difference between n = 1 (ground state) and n = 2, then the electron will drop from the excited state (n = 2) to the ground state (n = 1).
If the energy difference is equal to the energy level difference between n = 1 (ground state) and n = 3, then the electron will drop from the excited state (n = 3) to the ground state (n = 1).
If the energy difference is equal to the energy level difference between n = 1 (ground state) and n = 4, then the electron will drop from the excited state (n = 4) to the ground state (n = 1).
By comparing the energies of the absorbed and released photons, we can determine on which orbital the electron will drop.