Question

The energy of an electron in first Bohr orbit of H-atom is $-13.6$ eV. The magnitude of energy value of electron in the first excited state of Be$^{3+}$ is _____ eV (nearest integer value)

Hint:

- For hydrogen-like ions: \(E_n \propto Z^2/n^2\) - First excited state means \(n=2\) - Be\(^{3+}\) is isoelectronic with H but has \(Z=4\) - Energy becomes more negative (more stable) as \(Z\) increases

Answer 1

Correct Answer: 54

The total energy is given by:

\( E_T = - \frac{13.6 \, z^2}{n^2} \, \text{eV} \)

For the energy of the H atom in the 1st Bohr orbit, where \( z = 1 \) and \( n = 1 \):

\( E_1 = -13.6 \, \text{eV} \, [z = 1, n = 1] \)

For the \( \text{Be}^{3+} \) ion, the energy of the 1st excited state is given by \( z = 4 \) and \( n = 2 \):

\( \frac{E_H}{E_{\text{Be}^{3+}}} = \left( \frac{z_1}{n_1} \right)^2 \times \left( \frac{n_2}{z_2} \right)^2 \)

Substituting the values:

\( \frac{E_H}{E_{\text{Be}^{3+}}} = \frac{1^2}{1^2} \times \frac{2^2}{4^2} \)

Thus:

\( \frac{E_H}{E_{\text{Be}^{3+}}} = \frac{1}{1} \times \frac{4}{16} \)

So, the energy of the \( \text{Be}^{3+} \) ion is:

\( E_{\text{Be}^{3+}} = -13.6 \times 4 = -54.4 \, \text{eV} \)

Therefore, the magnitude of the energy is:

\( |E_{\text{Be}^{3+}}| = 54.4 \, \text{eV} \)

Answer 2

Step 1: Recall energy formula for hydrogen-like atoms The energy of an electron in the \(n^{th}\) orbit is given by: \[ E_n = -13.6 \frac{Z^2}{n^2} \text{ eV} \] where \(Z\) = atomic number, \(n\) = principal quantum number
Step 2: Identify parameters for Be\(^{3+}\) For Be\(^{3+}\): \[ Z = 4 \quad \text{(Beryllium)} \] First excited state corresponds to \(n = 2\)
Step 3: Calculate energy \[ E_2 = -13.6 \frac{4^2}{2^2} = -13.6 \times \frac{16}{4} = -13.6 \times 4 = -54.4 \text{ eV} \]
Step 4: Find magnitude \[ |E_2| = 54.4 \text{ eV} \approx 54 \text{ eV (nearest integer)} \]