A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of `10^(12)//sec`. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number `=6.02xx10^(23)" gm "mole^(-1)`)

Options

A. `5.5 N//m`
B. `6.4 N//m`
C. `7.1 N//m`
D. `2.2 N//m`

Answer

To find the force constant \( k \) of the bonds connecting silver atoms in a solid, we can use the relationship between frequency, mass, and the force constant in simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Frequency \( f = 10^{12} \, \text{s}^{-1} \)
- Molar mass of silver \( M = 108 \, \text{g/mol} \)
- Avogadro's number \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)

2. **Convert Molar Mass to Mass of One Atom**:
The mass \( m \) of one silver atom can be calculated using:
\[ m = \frac{M}{N_A} \]
Substituting the values:
\[ m = \frac{108 \, \text{g/mol}}{6.02 \times 10^{23} \, \text{mol}^{-1}} = \frac{108 \times 10^{-3} \, \text{kg}}{6.02 \times 10^{23}} \approx 1.79 \times 10^{-25} \, \text{kg} \]

3. **Relate Frequency to Force Constant**:
The frequency of oscillation in SHM is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Rearranging this formula to solve for \( k \):
\[ k = (2\pi f)^2 m \]

4. **Substitute the Values**:
Now substitute \( f \) and \( m \) into the equation:
\[ k = (2\pi \times 10^{12})^2 \times (1.79 \times 10^{-25}) \]

5. **Calculate \( k \)**:
First, calculate \( (2\pi \times 10^{12})^2 \):
\[ (2\pi \times 10^{12})^2 \approx (6.2832 \times 10^{12})^2 \approx 39.478 \times 10^{24} \approx 3.9478 \times 10^{25} \]
Now, multiply this by \( m \):
\[ k \approx 3.9478 \times 10^{25} \times 1.79 \times 10^{-25} \approx 7.06 \, \text{N/m} \]

6. **Final Result**:
Therefore, the force constant \( k \) is approximately:
\[ k \approx 7.1 \, \text{N/m} \]