CUET · PHYSICS · PYQ PAPER 2023
What is the total energy of electron in the \( n^{th} \) stationary orbit of hydrogen atom ?
- A \( \frac{13.6}{n^2} \text{ J} \)
- B \( -\frac{13.6}{n^2} \text{ eV} \)
- C \( 0.53n^2 \text{ J} \)
- D \( \frac{0.53}{n^2} \text{ eV} \)
Answer & Solution
Correct Answer
(B) \( -\frac{13.6}{n^2} \text{ eV} \)
Step-by-step Solution
Detailed explanation
According to Bohr's model, the energy of an electron is quantized and given by the formula \(E_n=-13.6 \frac{Z^2}{n^2} eV\). For a hydrogen atom \((Z=1)\), this simplifies to \(-\frac{13.6}{n^2} eV\), where the negative sign indicates that the electron is bound to the nucleus.…
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