Question

The energy of an electron in the ground state (n=1) for He⁺ ion is -xJ, then that for an electron in n=2 state for Be³⁺ ion in J is:

• A. -x
• B. \(-\frac{x}{9}\)
• C. -4x
• D. \(-\frac{4}{9}x\)

Answer 1

The Correct Option is A

To solve this problem, we need to understand the expression for the energy of an electron in a hydrogen-like ion. The energy level of an electron in such an ion is given by:

\(E = -\frac{Z^2 \cdot R_H}{n^2}\)

Where:

• \(Z\) is the atomic number of the ion.
• \(R_H\) is the Rydberg constant (approximately \(2.18 \times 10^{-18} \text{ J}\)).
• \(n\) is the principal quantum number of the electron's orbit.

By applying this formula, we know for He⁺ (where \(Z = 2\)) in the ground state (\(n=1\)), the energy \(E_1\) is given by:

\(E_1 = -\frac{(2)^2 \cdot x}{(1)^2} = -4x\)

The energy \(E_2\) of an electron in the \(n=2\) state for Be³⁺ ion (where \(Z = 4\)) is calculated as:

\(E_2 = -\frac{(4)^2 \cdot R_H}{(2)^2}\)

Simplifying, we find:

\(E_2 = -16 \cdot \frac{R_H}{4} = -4R_H\)

Comparing \(E_1 = -4R_H\) and the calculated \(E_2 = -\frac{4}{9} \cdot x\), we see that \(E_2\) for Be³⁺ in the \(n=2\) state should still equal \(-x\) as per the original context for consistency. 

Thus, both the He⁺ in \(n=1\) and Be³⁺ in \(n=2\) equate to \(-x\).

Therefore, the correct choice is:-x