A molecule of a substance has a permanent electric dipole moment of magnitude `10^(-29) `C m. A mole of this substance is polarized at low temperature by appling a strong elecrostatic field of magnitude `10^(6) V m^(-1)`. The direction of the field is suddenly changed by an angle of `60^(@)`. Estimate the heat released by the substance in aligning its dipole along the new direction of the field. For simplicity, assume `100%` polarisation of sample.

Options

A. `-6J`
B. `-3J`
C. `3J`
D. `6J`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the total dipole moment of one mole of the substance.
Given:
- Dipole moment of one molecule, \( p = 10^{-29} \, \text{C m} \)
- Number of molecules in one mole, \( N_A = 6 \times 10^{23} \)

The total dipole moment \( P \) for one mole is given by:
\[ P = N_A \times p = (6 \times 10^{23}) \times (10^{-29}) = 6 \times 10^{-6} \, \text{C m} \]

### Step 2: Calculate the initial potential energy \( U_i \).
The formula for potential energy \( U \) of a dipole in an electric field is:
\[ U = -P E \cos \theta \]
For the initial alignment, \( \theta = 0^\circ \) (parallel alignment). Thus:
\[ U_i = -P E \cos(0) = -P E \]
Substituting the values:
\[ U_i = - (6 \times 10^{-6}) \times (10^{6}) \times 1 = -6 \, \text{J} \]

### Step 3: Calculate the final potential energy \( U_f \).
Now, the field is rotated by \( 60^\circ \):
\[ U_f = -P E \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ U_f = - (6 \times 10^{-6}) \times (10^{6}) \times \frac{1}{2} = -3 \, \text{J} \]

### Step 4: Calculate the change in potential energy \( \Delta U \).
The heat released \( Q \) is equal to the change in potential energy:
\[ \Delta U = U_f - U_i \]
Substituting the calculated values:
\[ \Delta U = -3 - (-6) = -3 + 6 = 3 \, \text{J} \]

### Conclusion
The heat released by the substance in aligning its dipole along the new direction of the field is:
\[ \text{Heat released} = 3 \, \text{J} \]

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