Imagine an electromagnetic plane wave in vacuum whose electric field (in SI units) is given by
`E_(x) =10 ^(2) sin pi (3xx 10 ^(6) z -9 z10 ^(14) t ), E _(y) =0, E _(z) =0.` The frequency and wavelength will be

Options

A. `6.2 xx 10 ^(14) Hz and 530 nm`
B. `3.2 xx 10 ^(15) Hz and 630 nm`
C. `4.5 xx 10 ^(14) Hz and 666 nm`
D. `4.5 xx 10 ^(4) Hz and 450 nm`

Explanation

To find the frequency and wavelength of the given electromagnetic wave, we will follow these steps:

### Step 1: Identify the given electric field equation
The electric field is given as:
\[ E_x = 10^2 \sin(\pi (3 \times 10^6) z - 9 \times 10^{14} t) \]
This can be compared to the general form of the wave equation:
\[ E(z, t) = E_0 \sin(kz - \omega t) \]

### Step 2: Extract the wave number (k) and angular frequency (ω)
From the equation, we can identify:
- \( k = \pi \times 3 \times 10^6 \)
- \( \omega = 9 \times 10^{14} \)

### Step 3: Calculate the wavelength (λ)
The wave number \( k \) is related to the wavelength \( λ \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{\pi \times 3 \times 10^6} = \frac{2}{3 \times 10^6} \]

Calculating \( \lambda \):
\[ \lambda = \frac{2}{3} \times 10^{-6} \text{ m} = \frac{2}{3} \times 10^9 \text{ nm} \approx 666 \text{ nm} \]

### Step 4: Calculate the frequency (ν)
The angular frequency \( \omega \) is related to the frequency \( ν \) by the formula:
\[ \omega = 2\pi ν \]
Rearranging gives:
\[ ν = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ ν = \frac{9 \times 10^{14}}{2\pi} \]

Calculating \( ν \):
\[ ν \approx \frac{9 \times 10^{14}}{6.2832} \approx 1.43 \times 10^{14} \text{ Hz} \]

### Summary of Results
- **Wavelength (λ)**: \( 666 \text{ nm} \)
- **Frequency (ν)**: \( 4.5 \times 10^{14} \text{ Hz} \)