Question

What is the energy (in J) required to transfer the electron from \( n = 1 \) to \( n = 2 \) state in \( \text{Li}^{2+} \)? (K = constant \( 2.18 \times 10^{-18} \) J)

• A. \( \frac{4K}{27} \)
• B. \( 9K \)
• C. \( 8K \)
• D. \( \frac{27K}{4} \)

Hint:

- Use the formula \( E_n = - K \frac{Z^2}{n^2} \) for hydrogen-like atoms. - Always subtract \( E_1 \) from \( E_2 \) to get the excitation energy. - Ensure the atomic number \( Z \) is correctly substituted in calculations.

Answer 1

The Correct Option is D

Step 1: Energy Level Formula for Hydrogen-Like Atoms The energy of an electron in a hydrogen-like atom is given by: \[ E_n = - K \frac{Z^2}{n^2} \] where: - \( K = 2.18 \times 10^{-18} \) J (constant), - \( Z \) is the atomic number (for Lithium ion \( \text{Li}^{2+}, Z = 3 \)), - \( n \) is the principal quantum number. 

Step 2: Calculate Energy for \( n = 1 \) and \( n = 2 \) For \( n = 1 \): \[ E_1 = - K \frac{3^2}{1^2} = - 9K \] For \( n = 2 \): \[ E_2 = - K \frac{3^2}{2^2} = - K \frac{9}{4} = - \frac{9K}{4} \] 

Step 3: Calculate Energy Difference The energy required to excite the electron from \( n = 1 \) to \( n = 2 \) is: \[ \Delta E = E_2 - E_1 \] \[ \Delta E = \left( - \frac{9K}{4} \right) - (-9K) \] \[ \Delta E = - \frac{9K}{4} + 9K \] \[ \Delta E = 9K - \frac{9K}{4} = \frac{36K}{4} - \frac{9K}{4} = \frac{27K}{4} \]