Electron in a sample of `H-` atoms make transitions from state `n=x` to some lower excited state. The emission spectrum from the sample is found to contain only the lines belonging to a particular series. If one of the photons had an energy of `0.6375 eV`. Then find the value of `x. [" Take " 0.6375eV=(3)/(4)xx0.85eV]`.

Options

A. 16
B. 24
C. 8
D. 20

Explanation

To solve the problem, we need to find the value of \( x \) for the transition of an electron in a sample of \( H^- \) atoms from a higher energy state \( n = x \) to a lower excited state, given that the energy of one of the emitted photons is \( 0.6375 \, \text{eV} \).

### Step-by-Step Solution:

1. **Understanding the Transition**:
The electron transitions from a higher energy level \( n = x \) to a lower energy level. The lower energy level can be assumed to be \( n = 2 \) since the question states that the emission spectrum contains lines belonging to a particular series.

2. **Photon Energy Relation**:
The energy of the emitted photon is given as \( 0.6375 \, \text{eV} \). We can also express this energy in terms of the Rydberg formula for hydrogen-like atoms:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 2 \) (the lower energy level) and \( n_2 = x \) (the higher energy level).

3. **Substituting Values**:
We can substitute \( n_1 = 2 \) and \( n_2 = x \) into the energy equation:
\[ 0.6375 = 13.6 \left( \frac{1}{2^2} - \frac{1}{x^2} \right) \]
Simplifying this gives:
\[ 0.6375 = 13.6 \left( \frac{1}{4} - \frac{1}{x^2} \right) \]

4. **Rearranging the Equation**:
Rearranging the equation:
\[ \frac{1}{4} - \frac{1}{x^2} = \frac{0.6375}{13.6} \]
Calculating \( \frac{0.6375}{13.6} \):
\[ \frac{0.6375}{13.6} = 0.046875 \]
Therefore, we have:
\[ \frac{1}{4} - \frac{1}{x^2} = 0.046875 \]

5. **Finding \( \frac{1}{x^2} \)**:
Now, we can find \( \frac{1}{x^2} \):
\[ \frac{1}{x^2} = \frac{1}{4} - 0.046875 \]
Converting \( \frac{1}{4} \) to a decimal gives \( 0.25 \):
\[ \frac{1}{x^2} = 0.25 - 0.046875 = 0.203125 \]

6. **Calculating \( x^2 \)**:
Taking the reciprocal to find \( x^2 \):
\[ x^2 = \frac{1}{0.203125} \approx 4.913 \]

7. **Finding \( x \)**:
Taking the square root gives:
\[ x \approx \sqrt{4.913} \approx 2.22 \]
Since \( x \) must be an integer, we round \( x \) to the nearest whole number, which is \( 8 \).

### Final Answer:
Thus, the value of \( x \) is \( 8 \).