Question

The energy gap between valence band and the conduction band for a given material is 6 eV. What would be the energy associated with the second excited state of Li?

• A. \( -2.18 \times 10^{-18} \, \text{J} \)
• B. \( -4.905 \times 10^{-18} \, \text{J} \)
• C. \( -0.242 \times 10^{-18} \, \text{J} \)
• D. \( -3.26 \times 10^{-18} \, \text{J} \)

Hint:

For calculating energy transitions in atoms, the formula \(E_n = - \frac{13.6 \, \text{eV}}{n^2}\) can be used for hydrogen-like atoms to find the energy associated with any quantum state transition.

Answer 1

The Correct Option is A

The energy of the electron in the hydrogen atom is given by the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] where \(n\) is the principal quantum number. For Li, the ionization energy for the first electron transition is given. The energy associated with the second excited state of Li corresponds to the energy required to move from the first energy level \(n = 2\) to the second excited state \(n = 3\). Now using the above formula, we calculate the energy for the second excited state: \[ E_3 = - \frac{13.6 \, \text{eV}}{3^2} = - \frac{13.6}{9} = - 1.51 \, \text{eV} \] Now converting this to Joules: \[ E_3 = -1.51 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = -2.18 \times 10^{-18} \, \text{J} \] Thus, the energy associated with the second excited state of Li is \( -2.18 \times 10^{-18} \, \text{J} \).