The energy of an electron moving in `n^(th)` Bohr's orbit of an element is given by `E_(n)=(-13.6)/n^(2)Z^(2)` eV/ atom (Z=atomic number). The graph of E vs. `Z^(2)` (keeping "n" constant) will be :

Options

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D.

Answer

To solve the problem, we need to analyze the relationship between the energy of an electron in the nth Bohr orbit and the atomic number (Z) as given by the formula:

\[ E_n = \frac{-13.6}{n^2} Z^2 \]

### Step-by-Step Solution:

1. **Identify the relationship**:
The formula indicates that the energy \( E_n \) is proportional to \( Z^2 \). This can be expressed as:
\[ E_n \propto -Z^2 \]
This means that as \( Z^2 \) increases, \( E_n \) becomes more negative.

2. **Rearranging the equation**:
We can rewrite the equation in terms of \( Z^2 \):
\[ E_n = \frac{-13.6}{n^2} Z^2 \]
Here, we can see that if we plot \( E_n \) on the y-axis and \( Z^2 \) on the x-axis, we can identify the slope of the line.

3. **Identifying the slope**:
The equation can be rearranged to the form:
\[ E_n = m \cdot Z^2 \]
where \( m = \frac{-13.6}{n^2} \). Since \( m \) is negative, this indicates that the slope of the line is negative.

4. **Graph characteristics**:
Since the relationship is linear and the slope is negative, the graph of \( E \) versus \( Z^2 \) will be a straight line that slopes downwards from left to right.

5. **Conclusion**:
Therefore, the graph of \( E \) versus \( Z^2 \) (keeping \( n \) constant) will be a straight line with a negative slope.

### Final Answer:
The correct graph of \( E \) versus \( Z^2 \) is a straight line with a negative slope.