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

The energy of an electron in a hydrogen-like atom is given by:
$E_n = -\frac{13.6Z^2}{n^2} \, eV$,
where $Z$ is the atomic number and $n$ is the energy level.
For
$He^+$, $Z = 2$, and for 
$Be^{3+}$, $Z = 4$.
The ratio of energies in the ground state and $n = 2$ for $Be^{3+}$ gives:
$\frac{E}{E} = \frac{x}{1} = -x$.

Answer 2

Energy of an Electron in a Hydrogen-like Atom or Ion 

The energy of an electron in a hydrogen-like atom or ion is given by the formula:

$$ E_n = -\frac{Z^2}{n^2} \times 13.6 \text{ eV} $$

• \( E_n \) = Energy of the electron in the \( n \)-th energy level.
• \( Z \) = Atomic number of the ion.
• \( n \) = Principal quantum number.

Energy Calculation for \( He^+ \) Ion

For \( He^+ \), the atomic number \( Z = 2 \), and the energy for the ground state (\( n = 1 \)) is:

$$ E_1 = -\frac{2^2}{1^2} \times 13.6 \text{ eV} $$

$$ E_1 = -4 \times 13.6 \text{ eV} = -54.4 \text{ eV} $$

This is given as \( -x \) J, where:

$$ x = 54.4 \text{ eV} $$

Energy Calculation for \( Be^{3+} \) Ion

For \( Be^{3+} \), the atomic number \( Z = 4 \). The energy for an electron in the \( n = 2 \) state is:

$$ E_2 = -\frac{4^2}{2^2} \times 13.6 \text{ eV} $$

$$ E_2 = -\frac{16}{4} \times 13.6 \text{ eV} $$

$$ E_2 = -4 \times 13.6 \text{ eV} = -54.4 \text{ eV} $$

Since \( x = 54.4 \) eV, the energy for the \( n = 2 \) state for \( Be^{3+} \) is:

$$ E_2 = -\frac{x}{9} $$

Conclusion

The energy of the electron for the given ions is:

• \( He^+ \): \( E_1 = -x \)
• \( Be^{3+} \): \( E_2 = -\frac{x}{9} \)