The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number, and 13.6 eV is the ionization energy of hydrogen.
First, calculate the energy for \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.5111 \, \text{eV} \]
Next, calculate the energy for \( n = 5 \):
\[ E_5 = -\frac{13.6 \, \text{eV}}{5^2} = -\frac{13.6 \, \text{eV}}{25} = -0.5440 \, \text{eV} \]
The energy required for the transition from \( n = 3 \) to \( n = 5 \) is the difference between the energies of these two levels:
\[ \Delta E = E_5 - E_3 = (-0.5440 \, \text{eV}) - (-1.5111 \, \text{eV}) = 0.9671 \, \text{eV} \]
The energy required for the electronic transition from \( n = 3 \) to \( n = 5 \) in the hydrogen atom is \(\boxed{0.9671 \, \text{eV}}\).
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