Question

The energy of an electron in the second orbit of a hydrogen atom is \( -3.4 \, \text{eV} \). What is the energy of the electron in the third orbit? (Given: Energy of an electron in the \( n \)-th orbit of hydrogen is \( E_n = -\frac{13.6}{n^2} \, \text{eV} \)).

• A. \( -1.51 \, \text{eV} \)
• B. \( -2.27 \, \text{eV} \)
• C. \( -3.4 \, \text{eV} \)
• D. \( -6.04 \, \text{eV} \)

Hint:

For hydrogen atom energy levels, use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \). The energy becomes less negative as \( n \) increases, indicating the electron is farther from the nucleus.

Answer 1

The Correct Option is A

The energy of an electron in the \( n \)-th orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] For the third orbit (\( n = 3 \)):
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \, \text{eV} \] \[ E_3 \approx -\frac{13.6}{9} \approx -1.511 \, \text{eV} \] Rounding to two decimal places:
\[ E_3 \approx -1.51 \, \text{eV} \] To verify, note that the energy in the second orbit (\( n = 2 \)) is:
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \] This matches the given value, confirming the formula’s consistency. The energy becomes less negative (higher energy) as the orbit number increases, so the third orbit has a higher energy than the second.
Thus, the energy of the electron in the third orbit is \( -1.51 \, \text{eV} \).