Question

The radius of fourth orbit in \( He^+ \) ion is \( R_1 \) pm and radius of third orbit in \( Li^{2+} \) ion is \( R_2 \) pm. The value of \( (R_1 - R_2) \) in pm is:

• A. \( 132.25 \)
• B. \( 529.00 \)
• C. \( 264.50 \)
• D. \( 793.50 \)

Hint:

For hydrogenic atoms, use: \[ r_n = \frac{n^2 r_0}{Z} \] where \( r_0 \) is the Bohr radius, \( n \) is the orbit number, and \( Z \) is the atomic number.

Answer 1

The Correct Option is C

Step 1: Bohr Radius Formula The radius of the \( n \)th orbit for a hydrogen-like atom is given by: \[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z} \] or simplified as: \[ r_n = \frac{n^2 r_0}{Z} \] where: - \( r_0 \approx 53 \) pm (Bohr radius), - \( Z \) is the atomic number, - \( n \) is the principal quantum number. Step 2: Computing \( R_1 \) for \( He^+ \) For \( He^+ \), \( Z = 2 \), \( n = 4 \): \[ R_1 = \frac{4^2 \times 53}{2} \] \[ = \frac{16 \times 53}{2} \] \[ = \frac{848}{2} = 424 \text{ pm} \] Step 3: Computing \( R_2 \) for \( Li^{2+} \) For \( Li^{2+} \), \( Z = 3 \), \( n = 3 \): \[ R_2 = \frac{3^2 \times 53}{3} \] \[ = \frac{9 \times 53}{3} \] \[ = \frac{477}{3} = 159.5 \text{ pm} \] Step 4: Computing Difference \( R_1 - R_2 \) \[ R_1 - R_2 = 424 - 159.5 = 264.5 \text{ pm} \] Conclusion Thus, the correct answer is: \[ 264.50 \text{ pm} \]