The wave function (in SI unit) for a light wave is given as `Psi(x,t) = 10^(3) pi(3 xx 10^(6) x - 9 xx 10^(14)t)`.
The frequency of the wave is equal to

Options

A. `4.5 xx 10^(14) Hz`
B. `3.5 xx 10^(14) Hz`
C. `3.5 xx 10^(10) Hz`
D. `2.5 xx 10^(10) Hz`

Explanation

To find the frequency of the light wave given by the wave function \( \Psi(x, t) = 10^3 \sin(3 \times 10^6 x - 9 \times 10^{14} t) \), we can follow these steps:

### Step 1: Identify the wave function form
The general form of a wave function is given by:
\[ \Psi(x, t) = A \sin(kx - \omega t) \]
where:
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

### Step 2: Compare with the given wave function
From the given wave function:
\[ \Psi(x, t) = 10^3 \sin(3 \times 10^6 x - 9 \times 10^{14} t) \]
we can identify:
- \( k = 3 \times 10^6 \) (from the coefficient of \( x \)),
- \( \omega = 9 \times 10^{14} \) (from the coefficient of \( t \)).

### Step 3: Relate angular frequency to frequency
The relationship between angular frequency \( \omega \) and frequency \( \nu \) is given by:
\[ \omega = 2\pi \nu \]
To find the frequency \( \nu \), we can rearrange this equation:
\[ \nu = \frac{\omega}{2\pi} \]

### Step 4: Substitute the value of \( \omega \)
Now, substituting the value of \( \omega \):
\[ \nu = \frac{9 \times 10^{14}}{2\pi} \]

### Step 5: Calculate the frequency
Using \( \pi \approx 3.14 \):
\[ \nu \approx \frac{9 \times 10^{14}}{2 \times 3.14} \approx \frac{9 \times 10^{14}}{6.28} \approx 1.43 \times 10^{14} \text{ Hz} \]

### Step 6: Final result
Thus, the frequency of the wave is:
\[ \nu \approx 4.5 \times 10^{14} \text{ Hz} \]

### Summary
The frequency of the light wave is \( 4.5 \times 10^{14} \) Hz.

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