An element 'A' has face-centered cubic structure with edge length equal to 361pm .The apparent radius of atom 'A' is:

Options

A. `0.128` nm
B. `1.42` nm
C. `3.22` nm
D. `4.22` nm

Answer

To find the apparent radius of atom 'A' in a face-centered cubic (FCC) structure with a given edge length, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the relationship between edge length and radius in FCC:**
In a face-centered cubic structure, the relationship between the edge length (A) and the atomic radius (R) is given by the formula:
\[ A = 2\sqrt{2}R \]
This can also be rearranged to find R:
\[ R = \frac{A}{2\sqrt{2}} \]

2. **Substitute the given edge length:**
We are given that the edge length (A) is 361 pm (picometers). Plugging this value into the formula:
\[ R = \frac{361 \, \text{pm}}{2\sqrt{2}} \]

3. **Calculate the value of \( \sqrt{2} \):**
The value of \( \sqrt{2} \) is approximately 1.414. Therefore:
\[ R = \frac{361 \, \text{pm}}{2 \times 1.414} \]

4. **Perform the calculation:**
First, calculate \( 2 \times 1.414 \):
\[ 2 \times 1.414 \approx 2.828 \]
Now, divide 361 pm by 2.828:
\[ R \approx \frac{361}{2.828} \approx 127.6 \, \text{pm} \]

5. **Convert picometers to nanometers:**
Since 1 nm = 1000 pm, we convert the radius from picometers to nanometers:
\[ R \approx \frac{127.6 \, \text{pm}}{1000} = 0.1276 \, \text{nm} \]
Rounding this value gives:
\[ R \approx 0.128 \, \text{nm} \]

6. **Final answer:**
Thus, the apparent radius of atom 'A' is approximately **0.128 nm**.