Question

The energy of an electron in the ground state of Hydrogen atom is $-2.18 \times 10^{-18}$ J. What would be the energy associated with the second excited state of Li$^+$?

• A. $-2.18 \times 10^{-18}$ J
• B. $-4.905 \times 10^{-18}$ J
• C. $-0.242 \times 10^{-18}$ J
• D. $-3.26 \times 10^{-18}$ J

Hint:

The energy formula for hydrogen-like atoms is modified by the square of the atomic number, \(Z\), which helps us compute the energy for Li$^+$.

Answer 1

The Correct Option is A

The energy of an electron in a hydrogen atom is given by the equation: \[ E_n = - \frac{13.6 \text{ eV}}{n^2} \] where \(n\) is the principal quantum number. For lithium ion (Li$^+$), we can use the modified formula: \[ E_n = - \frac{13.6 \text{ eV}}{n^2} \times Z^2 \] where \(Z\) is the atomic number, which is 3 for Li$^+$. The second excited state corresponds to \(n = 3\). So, for Li$^+$, the energy associated with the second excited state (\(n = 3\)) is: \[ E_3 = - \frac{13.6 \times 3^2}{3^2} = - 2.18 \times 10^{-18} \, \text{J} \]