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 the formula for the energy of an electron in a hydrogen-like atom, which is given by:

\[ E_n = -\frac{mZ^2 e^4}{8 \epsilon_0 h^2 n^2} \]

where:
- \(E_n\) is the energy of the electron in the nth orbit,
- \(m\) is the mass of the electron,
- \(Z\) is the atomic number,
- \(e\) is the charge of the electron,
- \(\epsilon_0\) is the permittivity of free space,
- \(h\) is Planck's constant,
- \(n\) is the principal quantum number.

### Step 1: Identify the atomic numbers
For hydrogen (H), the atomic number \(Z_H = 1\).
For singly ionized helium (He\(^+\)), the atomic number \(Z_{He} = 2\).

### Step 2: Write the energy formula for both atoms
The energy for hydrogen in the nth orbit is:
\[ E_H = -\frac{m(1)^2 e^4}{8 \epsilon_0 h^2 n^2} = -\frac{m e^4}{8 \epsilon_0 h^2 n^2} \]

The energy for singly ionized helium in the nth orbit is:
\[ E_{He} = -\frac{m(2)^2 e^4}{8 \epsilon_0 h^2 n^2} = -\frac{4m e^4}{8 \epsilon_0 h^2 n^2} \]

### Step 3: Relate the energies
Now, we can express the energy of the helium atom in terms of the energy of the hydrogen atom:
\[ E_{He} = 4 \left(-\frac{m e^4}{8 \epsilon_0 h^2 n^2}\right) = 4 E_H \]

### Step 4: Substitute \(E_H\) with \(E_n\)
Since we know \(E_H = E_n\) (the energy of hydrogen in the nth orbit), we can write:
\[ E_{He} = 4 E_n \]

### Final Answer
Thus, the energy in the nth orbit of a singly ionized helium atom is:
\[ E_{He} = 4 E_n \]

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