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

The energy transition for an electron in a hydrogen-like atom is given by the formula: \(\Delta E = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\), where \(\Delta E\) is the change in energy, \(R_H\) is the Rydberg constant, \(\(n_1\) and \(\(n_2\) are the principal quantum numbers of the orbits involved in the transition.

Given that the electron transitions from the \(n\)th orbit to the \(n+1\) orbit with an energy change of \(1.47 × 10^{-1} J\), the equation becomes: \(1.47 × 10^{-1} = 2.18 × 10^{-18} \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)\).

Rearranging gives: \(\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{1.47 × 10^{-1}}{2.18 × 10^{-18}}\).

Calculate the right side: \(\frac{1.47 × 10^{-1}}{2.18 × 10^{-18}} ≈ 6.74311927 × 10^{16}\).

Thus, we have: \(\frac{1}{n^2} - \frac{1}{(n+1)^2} = 6.74311927 × 10^{16}\).

Let x = \frac{1}{n^2} and y = \frac{1}{(n+1)^2}, then x - y = 6.74311927 × 10^{16}\).

Rewriting y: y = \frac{1}{n^2} - \frac{2n+1}{n^2(n+1)^2}\. Solve for x: \frac{2n+1}{n^2(n+1)^2} = 6.74311927 × 10^{16}\), solve for n.

With simplification, solve n \approx 1.

Verify: The computed value of n is indeed 1, which is within the specified range 1, 1.