Consider the radiation emitted by large nu,ber of singly charged positive ions of a certain element. The sample emit fifteen types of spectral line. One of which is same as the first line of lyman series. What is the binding energy in the hinhest energy state of this configuration ?

Options

A. `13.6 eV`
B. `54.4 eV`
C. `10.2 eV`
D. `1.6 eV`

Explanation

To solve the problem, we need to find the binding energy in the highest energy state of a singly charged positive ion that emits 15 types of spectral lines, with one of them corresponding to the first line of the Lyman series.

### Step-by-Step Solution:

1. **Understanding the Number of Spectral Lines**:
The number of spectral lines emitted by an atom can be calculated using the formula:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
where \( n \) is the number of energy levels.

2. **Setting Up the Equation**:
We know from the problem that the number of spectral lines is 15. Therefore, we can set up the equation:
\[ \frac{n(n-1)}{2} = 15 \]
Multiplying both sides by 2 gives:
\[ n(n-1) = 30 \]

3. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ n^2 - n - 30 = 0 \]

4. **Factoring the Quadratic Equation**:
We can factor the quadratic equation:
\[ n^2 - n - 30 = (n - 6)(n + 5) = 0 \]
This gives us two solutions:
\[ n = 6 \quad \text{and} \quad n = -5 \]
Since \( n \) must be a positive integer, we take \( n = 6 \).

5. **Identifying the Element**:
The problem states that the ions are singly charged positive ions, which suggests that the atom behaves like a hydrogen-like atom. For \( n = 6 \), we can conclude that the ion is similar to a helium ion (He\(^+\)).

6. **Calculating the Binding Energy**:
The binding energy \( B \) for a hydrogen-like atom is given by the formula:
\[ B = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number. For helium, \( Z = 2 \) and \( n = 6 \):
\[ B = \frac{13.6 \cdot 2^2}{6^2} = \frac{13.6 \cdot 4}{36} \]

7. **Calculating the Final Value**:
Simplifying the expression:
\[ B = \frac{54.4}{36} \approx 1.511 \, \text{eV} \]
Rounding this gives approximately:
\[ B \approx 1.6 \, \text{eV} \]

8. **Conclusion**:
Thus, the binding energy in the highest energy state of this configuration is approximately \( 1.6 \, \text{eV} \).

### Final Answer:
The binding energy in the highest energy state of this configuration is \( 1.6 \, \text{eV} \).