If for a graph plotted between frequency of deflected X-ray beam and atomic number of the target metal `(sqrtv and Z)` gives a straight line with slope and intercept both as unity, then what will be the atomic number of the target metal if the observed frequency of X-rays is 2401 `sec^(-1)`?

Options

A. 47
B. 49
C. 50
D. 51

Explanation

To solve the problem, we will use Moseley’s law, which relates the frequency of X-rays emitted by an element to its atomic number. The relationship can be expressed as:

\[ \sqrt{v} = A \cdot Z - B \]

where:
- \( v \) is the frequency of the X-ray,
- \( Z \) is the atomic number of the target metal,
- \( A \) and \( B \) are constants.

Given that the graph of \(\sqrt{v}\) versus \(Z\) is a straight line with both slope and intercept equal to 1, we can set \( A = 1 \) and \( B = 1 \). Thus, the equation simplifies to:

\[ \sqrt{v} = Z - 1 \]

Now, we are given the observed frequency of X-rays as \( v = 2401 \, \text{sec}^{-1} \). We will substitute this value into the equation.

### Step-by-step Solution:

1. **Calculate \(\sqrt{v}\)**:
\[ \sqrt{v} = \sqrt{2401} \]
- Calculate \(\sqrt{2401}\):
\[ \sqrt{2401} = 49 \]

2. **Substitute \(\sqrt{v}\) into the equation**:
\[ 49 = Z - 1 \]

3. **Solve for \(Z\)**:
\[ Z = 49 + 1 \]
\[ Z = 50 \]

Thus, the atomic number of the target metal is **50**.

### Final Answer:
The atomic number of the target metal is **50**.