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 a target metal to its atomic number. The law can be expressed as:

\[ \sqrt{v} = aZ - b \]

where:
- \( v \) is the frequency of the emitted X-rays,
- \( Z \) 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 unity, we have:

- Slope \(a = 1\)
- Intercept \(b = 1\)

Now, substituting these values into the equation, we get:

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

Next, we are given the observed frequency of X-rays:

\[ v = 2401 \, \text{sec}^{-1} \]

Now, we need to find the atomic number \(Z\). First, we calculate \(\sqrt{v}\):

\[ \sqrt{v} = \sqrt{2401} \]

Calculating the square root:

\[ \sqrt{2401} = 49 \]

Now, substituting \(\sqrt{v}\) back into the equation:

\[ 49 = Z - 1 \]

To find \(Z\), we rearrange the equation:

\[ Z = 49 + 1 = 50 \]

Thus, the atomic number of the target metal is:

\[ \boxed{50} \]