What would be the change in energy for an atom when an electron goes from n=6 to n=3?

I'm assuming it would be negative because you would have to lose energy to go down energy levels. Would that be correct?

delta E = 2.180E-18 J*[1/(n1)^2 - 1/(n2)^2]

delta E = 2.180E-18 J*(1/3^2 - 1/6^2)
This is the amount of energy in J that is emitted when an electron moves from the n = 6 to n = 3. By the way this is for a H atom and not just any atom.

Yes, your assumption is correct. When an electron moves from a higher energy level to a lower energy level, it releases energy in the form of electromagnetic radiation. This release of energy is considered negative because the system (atom) loses energy.

To calculate the change in energy, you can use the equation:

ΔE = E_final - E_initial

In this case, E_final corresponds to the energy of the electron in the final energy level (n=3), and E_initial corresponds to the energy of the electron in the initial energy level (n=6).

The energy of an electron in a hydrogen-like atom is given by the equation:

E = -13.6 eV/n^2

where eV is electron volts and n is the principal quantum number.

For n = 6:
E_initial = -13.6 eV / (6^2)

For n = 3:
E_final = -13.6 eV / (3^2)

Substituting the values into the equation, we have:

ΔE = (-13.6 eV / 9) - (-13.6 eV / 36)

Simplifying this expression, we get:

ΔE ≈ -4.22 eV

Therefore, the change in energy for an electron going from n=6 to n=3 in an atom would be approximately -4.22 eV, indicating that the atom loses energy during this transition.