The momentum of a photon of an electromagnetic radiation is `3.3xx10^(-29)` kg m/sec. What is the frequency of the associated waves ? [`h=6.6xx10^(-34)` J-s]

Options

A. `3xx10^(10)` Hz
B. `1.5xx10^(13)`Hz
C. `4.5xx10^(13)`Hz
D. `6xx10^(10)Hz`

Explanation

To find the frequency of the associated waves of a photon given its momentum, we can use the relationship between momentum (p), frequency (f), and Planck's constant (h). The momentum of a photon is given by the formula:

\[ p = \frac{h}{\lambda} \]

where \( \lambda \) is the wavelength of the photon. We also know that the speed of light (c) is related to frequency and wavelength by:

\[ c = f \cdot \lambda \]

From these two equations, we can derive the frequency in terms of momentum:

1. **Express wavelength in terms of momentum**:
Rearranging the first equation gives us:
\[ \lambda = \frac{h}{p} \]

2. **Substitute wavelength into the speed of light equation**:
Substitute \( \lambda \) into the equation for the speed of light:
\[ c = f \cdot \left( \frac{h}{p} \right) \]

3. **Rearranging to find frequency**:
Rearranging the above equation to solve for frequency gives:
\[ f = \frac{p \cdot c}{h} \]

4. **Substituting the values**:
Now we can substitute the given values into the equation. We know:
- \( p = 3.3 \times 10^{-29} \, \text{kg m/s} \)
- \( h = 6.6 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

So, we have:
\[ f = \frac{(3.3 \times 10^{-29} \, \text{kg m/s}) \cdot (3 \times 10^8 \, \text{m/s})}{6.6 \times 10^{-34} \, \text{J s}} \]

5. **Calculating the frequency**:
Now, let's calculate:
\[ f = \frac{(3.3 \times 3) \times 10^{-29 + 8}}{6.6 \times 10^{-34}} \]
\[ f = \frac{9.9 \times 10^{-21}}{6.6 \times 10^{-34}} \]
\[ f = 1.5 \times 10^{13} \, \text{Hz} \]

Thus, the frequency of the associated waves is \( 1.5 \times 10^{13} \, \text{Hz} \).