Question

The energy of the hydrogen atom in its ground state (in eV) is \(-x\). The energy of \(He^+\) ion in its fourth orbit (in eV) is:

• A. \(+\dfrac{x}{5}\)
• B. \(-\dfrac{x}{2}\)
• C. \(+4x\)
• D. \(-\dfrac{x}{4}\)

Hint:

For any hydrogen-like species, \[ E_n=-\frac{13.6Z^2}{n^2}\ \text{eV} \] Always remember that energy is directly proportional to \(Z^2\) and inversely proportional to \(n^2\).

Answer 1

The Correct Option is D

Step 1: Energy of a hydrogen-like atom.
The energy of an electron in a hydrogen-like species is given by \[ E_n=-\frac{13.6Z^2}{n^2}\ \text{eV} \] where: \[ Z=\text{atomic number} \] and \[ n=\text{principal quantum number} \] For hydrogen, \[ Z=1,\qquad n=1 \] Hence, \[ E_1=-13.6\ \text{eV} \] According to the question, \[ E_1=-x \] Therefore, \[ x=13.6 \]

Step 2: Calculate the energy of \(He^+\) in the fourth orbit.
For the \(He^+\) ion, \[ Z=2 \] and for the fourth orbit, \[ n=4 \] Using the energy formula, \[ E_4=-\frac{13.6(2)^2}{(4)^2} \] \[ E_4=-\frac{13.6\times 4}{16} \] \[ E_4=-13.6\times \frac{1}{4} \] \[ E_4=-\frac{13.6}{4} \] Since \[ x=13.6, \] we get \[ E_4=-\frac{x}{4} \]

Step 3: Verification of options.
Comparing the obtained result \[ E_4=-\frac{x}{4} \] with the given options, we find that it matches option (4).

Step 4: Final conclusion.
Hence, the energy of the \(He^+\) ion in its fourth orbit is \[ \boxed{-\frac{x}{4}} \]