Question

According to Bohr's model, the highest kinetic energy is associated with the electron in the:

• A. First orbit of \(H\) atom
• B. First orbit of \(He^+\)
• C. Second orbit of \(He^+\)
• D. Second orbit of \(Li^{2+}\)

Hint:

For Bohr's model, remember that kinetic energy is proportional to \(\frac{Z^2}{n^2}\).

Answer 1

The Correct Option is B

The total energy (T.E.) of an electron in Bohr's \(n^\text{th}\) orbit is given by: \[ \text{T.E.} = -13.6 \frac{Z^2}{n^2} \, \text{eV/atom}. \] The kinetic energy (K.E.) of the electron is the negative of the total energy: \[ \text{K.E.} = -\text{T.E.} = 13.6 \frac{Z^2}{n^2}. \] Thus, K.E. is proportional to \(\frac{Z^2}{n^2}\). Calculate the K.E. for each option: 1. For the first orbit of \(H\) atom (\(n = 1, Z = 1\)): \[ \text{K.E.} \propto \frac{1^2}{1^2} = 1. \] 2. For the first orbit of \(He^+\) (\(n = 1, Z = 2\)): \[ \text{K.E.} \propto \frac{2^2}{1^2} = 4. \] 3. For the second orbit of \(He^+\) (\(n = 2, Z = 2\)): \[ \text{K.E.} \propto \frac{2^2}{2^2} = 1. \] 4. For the second orbit of \(Li^{2+}\) (\(n = 2, Z = 3\)): \[ \text{K.E.} \propto \frac{3^2}{2^2} = \frac{9}{4}. \] Comparing these values, the highest K.E. is associated with the first orbit of \(He^+\).