Question

The electron in the \(n^{th}\) orbit of \(Li^{2+}\) is excited to \((n + 1)\) orbit using the radiation of energy \(1.47 × 10^{–1}J \) (as shown in the diagrarn). The value of n is ______ .
Given: \(R_H = 2.18 × 10^{–18} J\).

Answer 1

Correct Answer: 1

Step 1: Using the formula for the energy difference between two orbits. 
The energy difference between two orbits for an electron in a hydrogen-like atom is given by the formula: \[ \Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( \Delta E \) is the energy difference, \( R_H \) is the Rydberg constant, \( Z \) is the atomic number (which is 3 for Li\(^{2+}\)), \( n_1 \) is the initial orbit, and \( n_2 \) is the final orbit. 
Step 2: Applying the given data. 
The electron in the nth orbit is excited to (n + 1) orbit using radiation of energy \( 1.47 \times 10^{-17} \, \text{J} \). Therefore, \( \Delta E = 1.47 \times 10^{-17} \, \text{J} \). We can substitute the known values into the formula: \[ 1.47 \times 10^{-17} = 2.18 \times 10^{-18} \times 9 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \] \[ \frac{1.47}{1.96} = \frac{3}{4} = \frac{1}{n^2} - \frac{1}{(n+1)^2} \] 
Step 3: Solving for \( n \). 
Solving the equation, we find that \( n = 1 \). Thus, the value of \( n \) is 1.