Question

A hydrogen atom falls from nth higher energy orbit to first energy orbit (n = 1). 

The energy released is equal to 12.75 eV. The nth orbit is

• A. \( n = 4 \)
• B. \( n = 3 \)
• C. \( n = 6 \)
• D. \( n = 5 \)

Hint:

In atomic transitions, the energy difference is inversely proportional to the square of the orbit numbers.

Answer 1

The Correct Option is A

Step 1: The energy released during the transition is given by the Rydberg formula: \[ \Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{eV} \] Step 2: Substituting \( n_1 = 1 \) and \( n_2 = n \), the energy released is: \[ \Delta E = 13.6 \left( 1 - \frac{1}{n^2} \right) \] Step 3: Given that \( \Delta E = 12.75 \) eV, we solve for \( n \): \[ 12.75 = 13.6 \left( 1 - \frac{1}{n^2} \right) \] \[ \frac{12.75}{13.6} = 1 - \frac{1}{n^2} \] \[ \frac{12.75}{13.6} = \frac{1}{n^2} \] Step 4: Solving for \( n \): \[ n^2 = \frac{13.6}{13.6 - 12.75} \quad \Rightarrow \quad n = 4 \] Thus, the correct answer is option (1), \( n = 4 \).