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.

Answer 2

To solve the problem, we need to determine which electron has the highest kinetic energy according to Bohr's model.

1. Kinetic energy in Bohr’s model:
For a hydrogen-like ion with atomic number \(Z\) and electron in the \(n^\text{th}\) orbit, the kinetic energy \(K_n\) is given by:

\[ K_n = \frac{Z^2 R_H}{n^2} \] where \(R_H\) is the Rydberg energy (~13.6 eV for hydrogen atom).

 

2. Calculate \(K_n\) for each option:

• First orbit of H atom (Z=1, n=1):
  \[ K = \frac{1^2 R_H}{1^2} = R_H = 13.6\, \text{eV} \]
• First orbit of He\(^+\) (Z=2, n=1):
  \[ K = \frac{2^2 R_H}{1^2} = 4 R_H = 54.4\, \text{eV} \]
• Second orbit of He\(^+\) (Z=2, n=2):
  \[ K = \frac{2^2 R_H}{2^2} = \frac{4 R_H}{4} = R_H = 13.6\, \text{eV} \]
• Second orbit of Li\(^{2+}\) (Z=3, n=2):
  \[ K = \frac{3^2 R_H}{2^2} = \frac{9 R_H}{4} = 2.25 R_H = 30.6\, \text{eV} \]

3. Comparing the kinetic energies:
\[ 54.4 > 30.6 > 13.6 = 13.6 \]

Final Answer:
The highest kinetic energy is for the electron in the first orbit of He\(^+\).