A particle of mass `9.1 xx 10^(-31)` kg travels in a medium with a speed of `10^(6)` m/s and a photon of a radiation of linear momentum `10^(-27)` kg m/s travels in vacuum. The wavelength of photon is _______ times the wavelength of the particle.

Answer

To solve the problem, we need to find the ratio of the wavelengths of a particle and a photon. We will use the de Broglie wavelength formula for the particle and the formula for the wavelength of a photon.

### Step-by-Step Solution:

1. **Calculate the wavelength of the particle (λ₁)**:
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \) (mass of the particle)
- \( v = 10^6 \, \text{m/s} \) (velocity of the particle)

Plugging in the values:
\[ \lambda_1 = \frac{6.63 \times 10^{-34}}{(9.1 \times 10^{-31})(10^6)} \]

2. **Calculate the wavelength of the photon (λ₂)**:
The wavelength of a photon is given by:
\[ \lambda = \frac{h}{p} \]
where:
- \( p = 10^{-27} \, \text{kg m/s} \) (momentum of the photon)

Plugging in the values:
\[ \lambda_2 = \frac{6.63 \times 10^{-34}}{10^{-27}} \]

3. **Find the ratio of the wavelengths (λ₂/λ₁)**:
To find how many times the wavelength of the photon is compared to the wavelength of the particle, we calculate:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{h}{p}}{\frac{h}{mv}} = \frac{mv}{p} \]

Substituting the values:
\[ \frac{\lambda_2}{\lambda_1} = \frac{(9.1 \times 10^{-31})(10^6)}{10^{-27}} \]

4. **Simplify the expression**:
\[ \frac{\lambda_2}{\lambda_1} = \frac{9.1 \times 10^{-25}}{10^{-27}} = 9.1 \times 10^{2} = 910 \]

### Final Answer:
The wavelength of the photon is **910 times** the wavelength of the particle.