The energy of the first electron in helium will be

Options

A. `-13.6 eV`
B. `-54.4 eV`
C. `-5.44 eV`
D. zero.

Explanation

To find the energy of the first electron in helium, we can use the formula for the energy of an electron in a hydrogen-like atom:

\[ E = -\frac{13.6 \, Z^2}{n^2} \]

Where:
- \( E \) is the energy of the electron,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step 1: Identify the values of Z and n
For helium:
- The atomic number \( Z = 2 \) (since helium has 2 protons).
- The principal quantum number \( n = 1 \) (since we are considering the first electron).

### Step 2: Substitute the values into the formula
Now we substitute the values of \( Z \) and \( n \) into the formula:

\[ E = -\frac{13.6 \times (2)^2}{(1)^2} \]

### Step 3: Calculate \( Z^2 \) and \( n^2 \)
Calculate \( Z^2 \) and \( n^2 \):
- \( Z^2 = 2^2 = 4 \)
- \( n^2 = 1^2 = 1 \)

### Step 4: Substitute the squared values back into the formula
Now substitute these squared values back into the equation:

\[ E = -\frac{13.6 \times 4}{1} \]

### Step 5: Perform the multiplication
Calculate the multiplication:

\[ E = -54.4 \, \text{eV} \]

### Final Answer
The energy of the first electron in helium is:

\[ E = -54.4 \, \text{eV} \]

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