If the value of `E=-78.4 "kcal//mol"`, the order of the orbit in hydrogen atom is-

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the problem of determining the order of the orbit in a hydrogen atom given the energy \( E = -78.4 \, \text{kcal/mol} \), we will follow these steps:

### Step 1: Understand the Energy Formula
The energy of an electron in a hydrogen atom can be expressed using the formula:
\[ E = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
For hydrogen, the atomic number \( Z = 1 \), so the formula simplifies to:
\[ E = -\frac{13.6}{n^2} \, \text{eV} \]

### Step 2: Convert Energy Units
We need to convert the energy from kilocalories per mole to electron volts. The conversion factor is:
\[ 1 \, \text{kcal/mol} = 0.0434 \, \text{eV} \]
Thus, we can convert \( -78.4 \, \text{kcal/mol} \) to electron volts:
\[ E = -78.4 \, \text{kcal/mol} \times 0.0434 \, \text{eV/kcal/mol} = -3.404 \, \text{eV} \]

### Step 3: Set Up the Equation
Substituting the value of \( E \) into the energy formula, we have:
\[ -3.404 = -\frac{13.6}{n^2} \]

### Step 4: Solve for \( n^2 \)
Removing the negative signs and rearranging gives:
\[ 3.404 = \frac{13.6}{n^2} \]
Multiplying both sides by \( n^2 \) and then dividing by \( 3.404 \):
\[ n^2 = \frac{13.6}{3.404} \]

### Step 5: Calculate \( n^2 \)
Now, we compute \( n^2 \):
\[ n^2 = \frac{13.6}{3.404} \approx 4 \]

### Step 6: Find \( n \)
Taking the square root of both sides gives:
\[ n = \sqrt{4} = 2 \]

### Conclusion
The order of the orbit in the hydrogen atom is \( n = 2 \).