Which of the following cannot be the value of ionisation energy for a hydrogen atom ?

Options

A. 0.85 eV
B. 3.4 eV
C. 1.51 eV
D. 0.27 eV

Answer

To determine which of the given values cannot be the ionization energy for a hydrogen atom, we need to use the formula for ionization energy:

\[ \text{Ionization Energy} = -\frac{13.6 \, \text{eV}}{n^2} \]

where \( n \) is the principal quantum number that can take positive integer values (1, 2, 3, ...).

### Step-by-Step Solution:

1. **Understand the Formula**: The ionization energy for a hydrogen atom is given by the formula \( E = -\frac{13.6}{n^2} \). This means that the ionization energy is negative and inversely proportional to the square of the principal quantum number \( n \).

2. **Evaluate Each Option**:
- **Option A: 0.85 eV**
- Set up the equation:
\[ 0.85 = \frac{13.6}{n^2} \]
- Rearranging gives:
\[ n^2 = \frac{13.6}{0.85} \approx 16 \]
- Therefore, \( n = 4 \) (which is possible).

- **Option B: 3.4 eV**
- Set up the equation:
\[ 3.4 = \frac{13.6}{n^2} \]
- Rearranging gives:
\[ n^2 = \frac{13.6}{3.4} \approx 4 \]
- Therefore, \( n = 2 \) (which is possible).

- **Option C: 1.51 eV**
- Set up the equation:
\[ 1.51 = \frac{13.6}{n^2} \]
- Rearranging gives:
\[ n^2 = \frac{13.6}{1.51} \approx 9 \]
- Therefore, \( n = 3 \) (which is possible).

- **Option D: 0.27 eV**
- Set up the equation:
\[ 0.27 = \frac{13.6}{n^2} \]
- Rearranging gives:
\[ n^2 = \frac{13.6}{0.27} \approx 50.37 \]
- Therefore, \( n \approx 7.07 \) (which is not a valid integer).

3. **Conclusion**: The value that cannot be the ionization energy for a hydrogen atom is **Option D: 0.27 eV**.