Quantum mechanics in atoms question

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A hydrogen atom transitions from the 3d level (n=3) to the 4f level (n=4) upon absorbing a photon. The wavelength of the absorbed photon can be calculated using the formula wavelength = hc/delta E, where delta E is the energy difference between the two levels. The energy levels for hydrogen are given by E_i = -13.6/n^2, and the absolute value of the energy difference, |E4 - E3|, is used for the calculation. It is emphasized that the sign of the energy difference does not affect the result, as only the magnitude is relevant. Understanding these principles is crucial for determining the wavelength of the absorbed photon.
jazzchan
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The question is:
A hydrogen atom absorbs a photon and makes a transition from a 3d level to a 4f level. What is the wavelength of the absorbed photon ?

As i know 3d is n = 3, 4f is n = 4

and wavelength = hc/delta E

so the deltaE is E4 - E3 or E3 - E4 ??

thanks..
 
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I think that's right. You also need to know what the energy is (compared to ionization energy, for instance) at a given principle quantum number.
 
Just use the absolute value of E4 - E3, because all you are interested in is the AMOUNT of energy needed to make the transition, which is a positive value. So it makes no difference if you use E4 - E3 or E3 - E4, since the absolute value produces the same result.

Since this is hydrogen, then E_i = -13.6/n^2 (in eV).
 
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