Question

Identify the pair of species having the same energy from the following: (The number given in the bracket corresponds to the principal quantum number (n) in which the electron is present.)

• A. \( \text{H}(n=1), \text{Li}^{2+} (n=1) \)
• B. \( \text{Li}^{2+} (n=3), \text{Be}^{3+} (n=4) \)
• C. \( \text{He}^+ (n=1), \text{Li}^{2+} (n=3) \)
• D. \( \text{H}(n=3), \text{Li}^{2+} (n=2) \)

Hint:

When dealing with hydrogen-like atoms, remember that the energy levels only depend on the atomic number \( Z \) and the principal quantum number \( n \).

Answer 1

The Correct Option is B

To identify the pair of species having the same energy, we need to consider the energy levels of electrons in hydrogen-like species. The energy of an electron in a hydrogen-like species is given by: \[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \] where: - \( Z \) is the atomic number, - \( n \) is the principal quantum number. Analysis of Each Option: 1. Option (1): \( \text{H}(n=1), \text{Li}^{2+} (n=1) \) - For \( \text{H} \) (Z = 1, n = 1): \[ E_1 = -\frac{13.6 \cdot 1^2}{1^2} = -13.6 \, \text{eV} \] - For \( \text{Li}^{2+} \) (Z = 3, n = 1): \[ E_1 = -\frac{13.6 \cdot 3^2}{1^2} = -122.4 \, \text{eV} \] - Energies are not the same. 2. Option (2): \( \text{Li}^{2+} (n=3), \text{Be}^{3+} (n=4) \) - For \( \text{Li}^{2+} \) (Z = 3, n = 3): \[ E_3 = -\frac{13.6 \cdot 3^2}{3^2} = -13.6 \, \text{eV} \] - For \( \text{Be}^{3+} \) (Z = 4, n = 4): \[ E_4 = -\frac{13.6 \cdot 4^2}{4^2} = -13.6 \, \text{eV} \] - Energies are the same. 3. Option (3): \( \text{He}^+ (n=1), \text{Li}^{2+} (n=3) \) - For \( \text{He}^+ \) (Z = 2, n = 1): \[ E_1 = -\frac{13.6 \cdot 2^2}{1^2} = -54.4 \, \text{eV} \] - For \( \text{Li}^{2+} \) (Z = 3, n = 3): \[ E_3 = -\frac{13.6 \cdot 3^2}{3^2} = -13.6 \, \text{eV} \] - Energies are not the same. 4. Option (4): \( \text{H}(n=3), \text{Li}^{2+} (n=2) \) - For \( \text{H} \) (Z = 1, n = 3): \[ E_3 = -\frac{13.6 \cdot 1^2}{3^2} = -1.51 \, \text{eV} \] - For \( \text{Li}^{2+} \) (Z = 3, n = 2): \[ E_2 = -\frac{13.6 \cdot 3^2}{2^2} = -30.6 \, \text{eV} \] - Energies are not the same. Final Answer: - The pair of species having the same energy is: \[ \boxed{\text{Li}^{2+} (n=3), \text{Be}^{3+} (n=4)} \] This corresponds to option (2).