Question

The energy of second orbit of hydrogen atom is -5.45x10^-19 J. What is the energy of first orbit of Li²⁺ ion (in J)

• A. -1.962x10^-18
• B. -1.962x10^-17
• C. -3.924x10^-17
• D. -3.924x10^-18

Answer 1

The Correct Option is B

To solve the problem, we need to find the energy of the first orbit of a Li²⁺ ion, given the energy of the second orbit of a hydrogen atom.

1. Understanding the Energy Relation:
The energy of an electron in the nth orbit of a hydrogen-like atom is given by:
$E_n = -13.6 \, \text{eV} \cdot \frac{Z^2}{n^2}$
where:
- $Z$ is the atomic number
- $n$ is the orbit number

2. Given Information:
Energy of second orbit of hydrogen atom = $-5.45 \times 10^{-19}$ J
This corresponds to $E_2$ for $Z = 1$:
$E_2 = -13.6 \cdot \frac{1^2}{2^2} = -3.4 \, \text{eV}$

3. Finding the Value of 1 eV in Joules:
From the above:
$-3.4 \, \text{eV} = -5.45 \times 10^{-19} \, \text{J}$
So, $1 \, \text{eV} = \frac{5.45 \times 10^{-19}}{3.4} \approx 1.6029 \times 10^{-19} \, \text{J}$

4. Energy of First Orbit of Li²⁺:
For Li²⁺, $Z = 3$, $n = 1$
$E_1 = -13.6 \cdot \frac{Z^2}{1^2} = -13.6 \cdot 9 = -122.4 \, \text{eV}$

5. Convert Energy to Joules:
$E_1 = -122.4 \cdot 1.6029 \times 10^{-19} \, \text{J}$
$E_1 \approx -1.962 \times 10^{-17} \, \text{J}$

Final Answer:
The energy of the first orbit of Li²⁺ ion is $ -1.962 \times 10^{-17} \, \text{J} $.