Question:
The minimum excitation energy of an electron revolving in the first orbit of hydrogen is:
The minimum excitation energy of an electron revolving in the first orbit of hydrogen is:
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The minimum excitation energy is the energy required to excite an electron from the ground state to the first excited state. It is simply the energy difference between those two states.
Updated On: May 9, 2025
- \( 3.4 \, \text{eV} \)
- \( 8.5 \, \text{eV} \)
- \( 10.2 \, \text{eV} \)
- \( 13.6 \, \text{eV} \)
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The Correct Option is C
Solution and Explanation
1. The energy levels of an electron in a hydrogen atom are given by the formula:
\[
E_n = - \frac{13.6}{n^2} \, \text{eV}
\]
Where \( n \) is the principal quantum number.
2. For the first orbit (\( n = 1 \)), the energy is: \[ E_1 = - 13.6 \, \text{eV} \] 3. For the second orbit (\( n = 2 \)), the energy is: \[ E_2 = - \frac{13.6}{4} = - 3.4 \, \text{eV} \] 4. The minimum excitation energy is the energy difference between these two levels: \[ E_{\text{excitation}} = E_2 - E_1 = (-3.4 \, \text{eV}) - (-13.6 \, \text{eV}) = 10.2 \, \text{eV} \] 5. Therefore, the minimum excitation energy is \( 10.2 \, \text{eV} \), which corresponds to option (3).
2. For the first orbit (\( n = 1 \)), the energy is: \[ E_1 = - 13.6 \, \text{eV} \] 3. For the second orbit (\( n = 2 \)), the energy is: \[ E_2 = - \frac{13.6}{4} = - 3.4 \, \text{eV} \] 4. The minimum excitation energy is the energy difference between these two levels: \[ E_{\text{excitation}} = E_2 - E_1 = (-3.4 \, \text{eV}) - (-13.6 \, \text{eV}) = 10.2 \, \text{eV} \] 5. Therefore, the minimum excitation energy is \( 10.2 \, \text{eV} \), which corresponds to option (3).
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