Question

Energy and radius of first Bohr orbit of He$^+$ and Li$^2+$ are:
Given: $ R_H = 2.18 \times 10^{-18} \, \text{J}, a_0 = 52.9 \, \text{pm} $

• A. \( E_n (\text{Li}^{2+}) = -8.72 \times 10^{-18} \, \text{J}, r_n (\text{Li}^{2+}) = 26.4 \, \text{pm}, E_n (\text{He}^{+}) = -19.62 \times 10^{-18} \, \text{J}, r_n (\text{He}^{+}) = 9.6 \, \text{pm} \)
• B. \( E_n (\text{Li}^{2+}) = -19.62 \times 10^{-16} \, \text{J}, r_n (\text{Li}^{2+}) = 17.6 \, \text{pm}, E_n (\text{He}^{+}) = -8.72 \times 10^{-16} \, \text{J}, r_n (\text{He}^{+}) = 26.4 \, \text{pm} \)
• C. \( E_n (\text{Li}^{2+}) = -8.72 \times 10^{-16} \, \text{J}, r_n (\text{Li}^{2+}) = 17.6 \, \text{pm}, E_n (\text{He}^{+}) = -19.62 \times 10^{-16} \, \text{J}, r_n (\text{He}^{+}) = 26.4 \, \text{pm} \)
• D. \( E_n (\text{Li}^{2+}) = -19.62 \times 10^{-18} \, \text{J}, r_n (\text{Li}^{2+}) = 17.5 \, \text{pm}, E_n (\text{He}^{+}) = -8.72 \times 10^{-18} \, \text{J}, r_n (\text{He}^{+}) = 26.4 \, \text{pm} \)

Hint:

For hydrogen-like ions, the energy and radius of the first Bohr orbit depend on the atomic number \( Z \). The energy is proportional to \( Z^2 \), and the radius is inversely proportional to \( Z \).

Answer 1

The Correct Option is A

The energy and radius of the first Bohr orbit are determined by the following equations:

1. Energy of the nth orbit for a hydrogen-like atom:

   \(E_n = - \frac{R_H}{n^2} Z^2\)

   Where \( Z \) represents the atomic number of the ion.

2. Radius of the nth orbit:

   \(r_n = \frac{a_0}{Z} \cdot n\)

Calculations are performed for \( n = 1 \) (the first Bohr orbit) for Li²⁺ (Z = 3) and He⁺ (Z = 2).

• For Li²⁺ (Z = 3):

  \(E_n (\text{Li}^{2+}) = - \frac{2.18 \times 10^{-18}}{1^2} \cdot 3^2 = -19.62 \times 10^{-18} \, \text{J}\)

  \(r_n (\text{Li}^{2+}) = \frac{52.9 \, \text{pm}}{3} = 17.6 \, \text{pm}\)

• For He⁺ (Z = 2):

  \(E_n (\text{He}^{+}) = - \frac{2.18 \times 10^{-18}}{1^2} \cdot 2^2 = -8.72 \times 10^{-18} \, \text{J}\)

  \(r_n (\text{He}^{+}) = \frac{52.9 \, \text{pm}}{2} = 26.4 \, \text{pm}\)

The calculated values are as follows:

• \( E_n (\text{Li}^{2+}) = -19.62 \times 10^{-18} \, \text{J}, r_n (\text{Li}^{2+}) = 17.6 \, \text{pm} \)
• \( E_n (\text{He}^{+}) = -8.72 \times 10^{-18} \, \text{J}, r_n (\text{He}^{+}) = 26.4 \, \text{pm} \)

Consequently, the correct response is (1).