Question

If the radius of ground state hydrogen is 51 pm, find out the radius of 5th orbit of Li2+ ions. (closest integer)

Answer 1

The radius of an orbit in a hydrogen-like atom is given by the formula: \[ r_n = \frac{n^2 \cdot r_1}{Z} \] where: - \( r_n \) is the radius of the \(n\)-th orbit, - \( r_1 \) is the radius of the ground state orbit (1st orbit) in hydrogen, - \( n \) is the principal quantum number (for the 5th orbit, \( n = 5 \)), - \( Z \) is the atomic number (for \( \text{Li}^{2+} \), \( Z = 3 \)). 
Step 1: Substitute the given values into the formula.
We are given that the radius of the ground state hydrogen orbit is 51 pm (picometers), and for \( \text{Li}^{2+} \), \( Z = 3 \), and \( n = 5 \): \[ r_5 = \frac{5^2 \cdot 51 \text{ pm}}{3} \] Step 2: Calculate the radius of the 5th orbit of \( \text{Li}^{2+} \).
\[ r_5 = \frac{25 \cdot 51}{3} = \frac{1275}{3} = 425 \text{ pm} \] Final Answer:
The radius of the 5th orbit of \( \text{Li}^{2+} \) is 425 pm.