KCET · Physics · Atomic Physics
The radius of first orbit in hydrogen atom is \(5.3 \times 10^{-11}\) m. The kinetic energy \(E_K\), potential energy \(E_P\) and total energy \(E_T\) of electron in first orbit are
- A \(E_K = -13.6\text{ eV}, E_P = 27.2\text{ eV}, E_T = 13.6\text{ eV}\)
- B \(E_K = 13.6\text{ eV}, E_P = -27.2\text{ eV}, E_T = -13.6\text{ eV}\)
- C \(E_K = -27.2\text{ eV}, E_P = -13.6\text{ eV}, E_T = 13.6\text{ eV}\)
- D \(E_K = 13.6\text{ eV}, E_P = -6.8\text{ eV}, E_T = -13.6\text{ eV}\)
Answer & Solution
Correct Answer
(B) \(E_K = 13.6\text{ eV}, E_P = -27.2\text{ eV}, E_T = -13.6\text{ eV}\)
Step-by-step Solution
Detailed explanation
The total energy of an electron in the first orbit of a hydrogen atom is known to be \(E_T = -13.6 \text{ eV}\).
The relationship between kinetic energy \(E_K\), potential energy \(E_P\), and total energy \(E_T\) for an electron in a hydrogen atom is given by:
\(E_K = -E_T\)
\(E_P = 2E_T\)
Substituting the value of \(E_T\) into these relations:
\(E_K = -(-13.6 \text{ eV}) = 13.6 \text{ eV}\)
\(E_P = 2(-13.6 \text{ eV}) = -27.2 \text{ eV}\)
Thus, the energies are \(E_K = 13.6 \text{ eV}\), \(E_P = -27.2 \text{ eV}\), and \(E_T = -13.6 \text{ eV}\).
Answer: \(E_K = 13.6\text{ eV}, E_P = -27.2\text{ eV}, E_T = -13.6\text{ eV}\)
The relationship between kinetic energy \(E_K\), potential energy \(E_P\), and total energy \(E_T\) for an electron in a hydrogen atom is given by:
\(E_K = -E_T\)
\(E_P = 2E_T\)
Substituting the value of \(E_T\) into these relations:
\(E_K = -(-13.6 \text{ eV}) = 13.6 \text{ eV}\)
\(E_P = 2(-13.6 \text{ eV}) = -27.2 \text{ eV}\)
Thus, the energies are \(E_K = 13.6 \text{ eV}\), \(E_P = -27.2 \text{ eV}\), and \(E_T = -13.6 \text{ eV}\).
Answer: \(E_K = 13.6\text{ eV}, E_P = -27.2\text{ eV}, E_T = -13.6\text{ eV}\)
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