The kinetic energy of a particle is equal to the energy of a photon. The particle maves at 5% of the speed of light . The ratio of the photon wavelength to the de Broglie wavelength of the particle is [No nee to use reletivistic formula for particle.]

Options

A. 40
B. 4
C. 2
D. 80

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Understand the relationship between kinetic energy and photon energy
We know that the kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
The energy of a photon is given by:
\[ E = \frac{hc}{\lambda} \]
According to the problem, these two energies are equal:
\[ \frac{1}{2} mv^2 = \frac{hc}{\lambda_{photon}} \]

### Step 2: Rearranging the equation for photon wavelength
From the above equation, we can rearrange it to find the wavelength of the photon:
\[ \lambda_{photon} = \frac{2hc}{mv^2} \]

### Step 3: Calculate the de Broglie wavelength of the particle
The de Broglie wavelength (\(\lambda_{de Broglie}\)) of a particle is given by:
\[ \lambda_{de Broglie} = \frac{h}{p} = \frac{h}{mv} \]

### Step 4: Find the ratio of the photon wavelength to the de Broglie wavelength
We need to find the ratio:
\[ \frac{\lambda_{photon}}{\lambda_{de Broglie}} = \frac{\frac{2hc}{mv^2}}{\frac{h}{mv}} = \frac{2hc}{mv^2} \cdot \frac{mv}{h} \]
This simplifies to:
\[ \frac{\lambda_{photon}}{\lambda_{de Broglie}} = \frac{2c}{v} \]

### Step 5: Substitute the given speed of the particle
The problem states that the particle moves at 5% of the speed of light, so:
\[ v = 0.05c = \frac{c}{20} \]
Now, substituting \(v\) into the ratio:
\[ \frac{\lambda_{photon}}{\lambda_{de Broglie}} = \frac{2c}{\frac{c}{20}} = 2c \cdot \frac{20}{c} = 40 \]

### Final Answer
Thus, the ratio of the photon wavelength to the de Broglie wavelength of the particle is:
\[ \boxed{40} \]