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 Li²⁺

Answer 1

The Correct Option is B

Kinetic Energy Calculation for Various Orbits 

The total energy (T.E.) of an electron in Bohr’s nth orbit is given by:

\[ T.E. = -\frac{13.6Z^2}{n^2} \, \text{eV/atom} \]

The kinetic energy (K.E.) of the electron is the negative of the total energy:

\[ K.E. = -T.E. = \frac{13.6Z^2}{n^2} \]

Thus, \( K.E. \) is proportional to \( \frac{Z^2}{n^2} \).

Calculate the K.E. for Each Option:

• For the first orbit of H atom (\( n = 1, Z = 1 \)): \[ K.E. \propto \frac{1^2}{1^2} = 1 \]
• For the first orbit of \( \text{He}^+ \) (\( n = 1, Z = 2 \)): \[ K.E. \propto \frac{2^2}{1^2} = 4 \]
• For the second orbit of \( \text{He}^+ \) (\( n = 2, Z = 2 \)): \[ K.E. \propto \frac{2^2}{2^2} = 1 \]
• For the second orbit of \( \text{Li}^{2+} \) (\( n = 2, Z = 3 \)): \[ K.E. \propto \frac{3^2}{2^2} = \frac{9}{4} \]

Conclusion:

Comparing these values, the highest K.E. is associated with the first orbit of \( \text{He}^+ \), with a value of 4.