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 calculated using the formula:

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

For the hydrogen atom in its ground state (1st Bohr orbit), with \( z = 1 \) and \( n = 1 \):

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

For the \( \text{Be}^{3+} \) ion, considering the first excited state, where \( z = 4 \) and \( n = 2 \):

The ratio of the energy of the hydrogen atom to the energy of \( \text{Be}^{3+} \) is given by:

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

Upon substitution of the values:

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

This simplifies to:

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

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

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

The absolute value of the energy for the \( \text{Be}^{3+} \) ion is:

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