Question

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 eV, \(\quad\) E\(_P\) = 27.2 eV, \(\quad\) E\(_T\) = 13.6 eV
• B. E\(_K\) = 13.6 eV, \(\quad\) E\(_P\) = - 27.2 eV, \(\quad\) E\(_T\) = - 13.6 eV
• C. E\(_K\) = - 27.2 eV, \(\quad\) E\(_P\) = - 13.6 eV, \(\quad\) E\(_T\) = 13.6 eV
• D. E\(_K\) = 13.6 eV, \(\quad\) E\(_P\) = - 6.8 eV, \(\quad\) E\(_T\) = - 13.6 eV

Hint:

Always remember: Kinetic energy is positive, Total energy is negative (for bound states), and Potential energy is twice the total energy (also negative). This immediately eliminates options (1) and (3).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the different energy components of an electron in the ground state (\(n=1\)) of a hydrogen atom.
Step 2: Key Formula or Approach:
For a Bohr atom, the relationship between energies is:
1. Total Energy \(E_T = \frac{E_P}{2} = -E_K\)
2. For Hydrogen ground state, \(E_T = -13.6 \text{ eV}\)
Step 3: Detailed Explanation:
In the first orbit of Hydrogen (\(n=1\)):
- Total Energy (\(E_T\)): By Bohr's formula, \(E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}\). For \(Z=1, n=1\), \(E_T = -13.6 \text{ eV}\).
- Kinetic Energy (\(E_K\)): Kinetic energy is always positive and \(E_K = -E_T = -(-13.6) = 13.6 \text{ eV}\).
- Potential Energy (\(E_P\)): Potential energy is negative for an attractive system and \(E_P = 2 E_T = 2 \times (-13.6) = -27.2 \text{ eV}\).
Checking the options, (2) perfectly matches these calculations.
Step 4: Final Answer:
The correct set of energies is given in option (2).