A target element A is bombarded with electrons and the wavelengths of the characterstic spectrum are measured. A second characteristic spectrum is also obtained , because of an impurity in the target. The wavelengths of the `K_(alpha)` lines are 196 pm (element A) and 169 pm (impurity). If the atomic number of impurity is `z=(10lambda-1)` . Find the value of `lambda` . (atomic number of element A is 27).

Answer

To solve the problem, we will use Moseley's Law, which states that the wavelength of the characteristic X-ray spectrum is inversely proportional to the square of the atomic number of the element.

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Wavelength of K-alpha line for element A, \( \lambda_1 = 196 \) pm.
- Wavelength of K-alpha line for impurity, \( \lambda_2 = 169 \) pm.
- Atomic number of element A, \( Z_A = 27 \).
- Atomic number of impurity, \( Z = 10\lambda - 1 \).

2. **Apply Moseley's Law**:
According to Moseley's Law:
\[ \lambda \propto \frac{1}{(Z - 1)^2} \]
This can be expressed in terms of a constant \( k \):
\[ \lambda = \frac{k}{(Z - 1)^2} \]

3. **Set Up the Ratio**:
For the two elements, we can write:
\[ \frac{\lambda_1}{\lambda_2} = \frac{(Z_2 - 1)^2}{(Z_1 - 1)^2} \]
Substituting the values:
\[ \frac{196}{169} = \frac{(Z - 1)^2}{(27 - 1)^2} \]
Simplifying the right side:
\[ \frac{196}{169} = \frac{(Z - 1)^2}{26^2} \]

4. **Cross Multiply**:
\[ 196 \cdot 26^2 = 169 \cdot (Z - 1)^2 \]

5. **Calculate \( 26^2 \)**:
\[ 26^2 = 676 \]
Thus,
\[ 196 \cdot 676 = 169 \cdot (Z - 1)^2 \]

6. **Calculate \( 196 \cdot 676 \)**:
\[ 196 \cdot 676 = 132656 \]
So we have:
\[ 132656 = 169 \cdot (Z - 1)^2 \]

7. **Divide Both Sides by 169**:
\[ (Z - 1)^2 = \frac{132656}{169} \]
Calculating the right side:
\[ (Z - 1)^2 = 784 \]

8. **Take the Square Root**:
\[ Z - 1 = 28 \quad \text{or} \quad Z - 1 = -28 \quad (\text{not valid since } Z \text{ must be positive}) \]
Thus,
\[ Z = 28 + 1 = 29 \]

9. **Relate \( Z \) to \( \lambda \)**:
We know:
\[ Z = 10\lambda - 1 \]
Therefore:
\[ 29 = 10\lambda - 1 \]
Solving for \( \lambda \):
\[ 10\lambda = 30 \quad \Rightarrow \quad \lambda = 3 \]

### Final Answer:
The value of \( \lambda \) is \( 3 \).