Question

which of the following quantum leaps would be associated with the greatest energy of emitted light?
○ n = 2 to n = 5
○ n = 4 to n = 5
○ n = 5 to n = 1
○ n = 5 to n = 4

Explanation:

• When an electron makes a quantum leap (transition), energy is emitted when it moves from a higher - energy level (\(n_{high}\)) to a lower - energy level (\(n_{low}\)). The energy of the emitted photon is given by \(E = h

u=E_{high}-E_{low}\), where \(E =-\frac{13.6}{n^{2}}\text{ eV}\) (for hydrogen - like atoms).

• For \(n = 2\) to \(n = 5\), the electron is absorbing energy (not emitting).
• For \(n = 4\) to \(n = 5\), the electron is absorbing energy (not emitting).
• For \(n = 5\) to \(n = 1\):
• Calculate the energy difference. Using \(E =-\frac{13.6}{n^{2}}\text{ eV}\), \(E_{5}=-\frac{13.6}{5^{2}}=- 0.544\text{ eV}\) and \(E_{1}=-13.6\text{ eV}\). The energy of the emitted photon is \(E = E_{5}-E_{1}=-0.544-(-13.6)=13.056\text{ eV}\).
• For \(n = 5\) to \(n = 4\):
• \(E_{5}=-\frac{13.6}{5^{2}}=-0.544\text{ eV}\) and \(E_{4}=-\frac{13.6}{4^{2}}=- 0.85\text{ eV}\). The energy of the emitted photon is \(E = E_{5}-E_{4}=-0.544-(-0.85) = 0.306\text{ eV}\)

Answer:

\(n = 5\) to \(n = 1\)