Question

If the energy of a hydrogen atom in nth orbit is $E_n$, then energy in the nth orbit of a singly ionized helium atom will be

• A. $ 4E_{n} $
• B. $ E_{n}/4 $
• C. $ 2E_{n} $
• D. $ E_{n}/2 $

Answer 1

The Correct Option is A

To determine the energy of a singly ionized helium atom in the nth orbit, we start by understanding the formula for the energy levels of hydrogen-like atoms.

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

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

where Z is the atomic number, and n is the principal quantum number (orbit number).

For a hydrogen atom, Z = 1, so its energy in the nth orbit is:

E_{n,\text{hydrogen}} = -\frac{1^2 \times 13.6 \, \text{eV}}{n^2} = -\frac{13.6 \, \text{eV}}{n^2}

Given that the energy of the hydrogen atom in the nth orbit is E_n, this can be equated to:

E_n = -\frac{13.6 \, \text{eV}}{n^2}

For a singly ionized helium atom (He^+), the atomic number Z = 2. So, the energy in the nth orbit will be:

E_{n,\text{He}^+} = -\frac{2^2 \times 13.6 \, \text{eV}}{n^2} = -\frac{4 \times 13.6 \, \text{eV}}{n^2}

Therefore, the energy of the nth orbit of a singly ionized helium atom will be:

E_{n,\text{He}^+} = 4(-\frac{13.6 \, \text{eV}}{n^2}) = 4 E_n

Hence, the energy in the nth orbit of a singly ionized helium atom is 4E_n, which matches the given correct answer.

Conclusion: The correct answer is 4E_n. This is based on the insight that the energy levels are directly proportional to the square of the atomic number Z^2 in hydrogen-like atoms.