Question

The radius of fifth orbit of the Li++ is __________ × 10^-12 m.
Take : radius of hydrogen atom = 0.51 Å

Answer 1

Correct Answer: 425

\(𝑟_𝑛 = \frac{0.51𝑛^2}{𝑍} 𝐴_0 \)
For Li++, z=3.
So \(𝑟_5 = 0.51 × \frac{25}{3} × 10^{−10}\) 𝑚 = 17 × 25 × \(10^{−12}𝑚\) = 425 × \(10^{−12}\)m

Answer 2

Bohr's Model and Orbital Radius 

Step 1: Bohr’s Model and Orbital Radius

According to Bohr’s model, the radius of the n^th orbit of a hydrogen-like atom is given by:

\( r_n = \frac{0.51n^2}{Z} \, \text{Å} \)

where \( n \) is the principal quantum number and \( Z \) is the atomic number.

Step 2: Radius of Li⁺⁺ Fifth Orbit

For Li⁺⁺, \( Z = 3 \) and \( n = 5 \). Substituting these values into the formula:

\( r_5 = \frac{0.51 \times 5^2}{3} \, \text{Å} = \frac{0.51 \times 25}{3} \, \text{Å} \)

Now, convert the value to meters:

\( r_5 = 0.51 \times \frac{25}{3} \times 10^{-10} \, \text{m} = 17 \times 25 \times 10^{-12} \, \text{m} = 425 \times 10^{-12} \, \text{m} \)

Conclusion:

The radius of the fifth orbit of Li⁺⁺ is 425 × 10⁻¹² m.