Frequency of the X-rays emitted by an element is `100 s^(-1)`. If constant in the Moseley equation are `a = b = 1`, the element will be

Options

A. Na
B. K
C. Rb
D. Cs

Explanation

To solve the problem, we will use Moseley's equation, which relates the frequency of X-rays emitted by an element to its atomic number. The equation is given by:

\[ \nu = A(Z - B) \]

where:
- \( \nu \) is the frequency of the emitted X-rays,
- \( A \) and \( B \) are constants,
- \( Z \) is the atomic number of the element.

Given:
- Frequency \( \nu = 100 \, \text{s}^{-1} \)
- Constants \( A = 1 \) and \( B = 1 \)

**Step 1: Substitute the values into Moseley's equation.**

We start by substituting the known values into the equation:

\[ 100 = 1(Z - 1) \]

**Step 2: Simplify the equation.**

This simplifies to:

\[ 100 = Z - 1 \]

**Step 3: Solve for the atomic number \( Z \).**

Now, we can solve for \( Z \):

\[ Z = 100 + 1 \]
\[ Z = 101 \]

**Step 4: Identify the element with atomic number 101.**

The element with atomic number 101 is Mendelevium (Md).

**Final Answer:**
The element will be Mendelevium (Md).

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