If the energy of a hydrogen atom in `nth` orbit is `E_(n)`, then energy in the nth orbit of a singly ionised helium atom will be

Options

A. `4E_(n)`
B. `E_(n)/4`
C. `2E_(n)`
D. `E_(n)/2`

Answer

To find the energy of a singly ionized helium atom in the nth orbit, we can use Bohr's model of the atom. The energy of an electron in the nth orbit of an atom is given by the formula:

\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]

where:
- \( E_n \) is the energy in the nth orbit,
- \( Z \) is the atomic number of the atom,
- \( n \) is the principal quantum number (the orbit number).

### Step 1: Energy of Hydrogen Atom
For hydrogen (\( Z = 1 \)):
\[ E_H = -\frac{13.6 \times 1^2}{n^2} = -\frac{13.6}{n^2} \text{ eV} \]

### Step 2: Energy of Singly Ionized Helium Atom
For singly ionized helium (\( Z = 2 \)):
\[ E_{He} = -\frac{13.6 \times 2^2}{n^2} = -\frac{13.6 \times 4}{n^2} = -\frac{54.4}{n^2} \text{ eV} \]

### Step 3: Relate the Energies
Now, we can relate the energy of the singly ionized helium atom to that of the hydrogen atom:
\[ E_{He} = -\frac{54.4}{n^2} \text{ eV} \]
\[ E_H = -\frac{13.6}{n^2} \text{ eV} \]

### Step 4: Expressing \( E_{He} \) in terms of \( E_H \)
Now, we can express \( E_{He} \) in terms of \( E_H \):
\[ E_{He} = 4 \times E_H \]
Since \( E_H = -\frac{13.6}{n^2} \text{ eV} \), we have:
\[ E_{He} = 4 \left(-\frac{13.6}{n^2}\right) = -\frac{54.4}{n^2} \text{ eV} \]

### Conclusion
Thus, the energy in the nth orbit of a singly ionized helium atom is:
\[ E_{He} = 4 E_H \]
Since \( E_H = E_n \):
\[ E_{He} = 4 E_n \]

### Final Answer
The energy in the nth orbit of a singly ionized helium atom will be \( 4 E_n \).

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