Question

The energy of the second orbit of a hydrogen atom is \( -5.45 \times 10^{-19} \) J. What is the energy of the first orbit of \( Li^{2+} \) ion (in J)?

• A. \( -1.962 \times 10^{-18} \)
• B. \( -1.962 \times 10^{-17} \)
• C. \( -3.924 \times 10^{-17} \)
• D. \( -3.924 \times 10^{-18} \)

Hint:

The energy levels of multi-electron atoms scale with \( Z^2 \); for \( Li^{2+} \), energy is 9 times that of hydrogen.

Answer 1

The Correct Option is B

Step 1: {Using Energy Formula}
The energy of an electron in an orbit is given by: \[ E_n = -13.6 \frac{Z^2}{n^2} eV \] Step 2: {Finding Energy for \( Li^{2+} \)}
For \( Li^{2+} \), \( Z = 3 \) and for the first orbit \( n = 1 \): \[ E_1 = -13.6 \times \frac{3^2}{1^2} eV \] \[ E_1 = -122.4 eV \] Step 3: {Converting to Joules}
\[ E_1 = (-122.4) \times (1.602 \times 10^{-19}) \] \[ E_1 = -1.962 \times 10^{-17} { J} \] Thus, the correct answer is (B).

Answer 2

Step 1: Understand the energy level formula for hydrogen-like species
For a hydrogen-like ion, the energy of the nth orbit is given by:
\( E_n = - \frac{13.6 \times Z^2}{n^2} \) eV
Or in joules:
\( E_n = - \frac{2.18 \times 10^{-18} \times Z^2}{n^2} \) J
Where:
- \( Z \) is the atomic number
- \( n \) is the principal quantum number (orbit number)

Step 2: Use the energy of the second orbit of hydrogen
Given: Energy of 2nd orbit of hydrogen atom = \( -5.45 \times 10^{-19} \) J
We can generalize this using the energy level formula:
Hydrogen has \( Z = 1 \), so:
\( E_2(H) = - \frac{2.18 \times 10^{-18} \times 1^2}{2^2} = - \frac{2.18 \times 10^{-18}}{4} \approx -5.45 \times 10^{-19} \) J
This confirms the formula's accuracy.

Step 3: Calculate the energy for the 1st orbit of \( Li^{2+} \)
For \( Li^{2+} \), \( Z = 3 \) and \( n = 1 \)
Use the formula:
\( E_1(Li^{2+}) = - \frac{2.18 \times 10^{-18} \times 3^2}{1^2} \)
\( E_1 = -2.18 \times 10^{-18} \times 9 = -1.962 \times 10^{-17} \) J

Step 4: Final Answer
The energy of the first orbit of \( Li^{2+} \) is:
\( -1.962 \times 10^{-17} \) J