A particle of mass $$9.1 \times 10^{-31}$$ kg travels in a medium with a speed of $$10^6$$ m s$$^{-1}$$ and a photon of radiation of linear momentum $$10^{-27}$$ kg m s$$^{-1}$$ travels in a vacuum. The wavelength of the photon is _________ times the wavelength of the particle.

For a material particle, the de-Broglie relation gives the wavelength as $$\lambda = \frac{h}{p}$$, where $$h$$ is Planck’s constant and $$p$$ is the linear momentum.

Case 1: Particle of mass $$m = 9.1 \times 10^{-31}\,\text{kg}$$ moving with speed $$v = 10^{6}\,\text{m s}^{-1}$$
  Momentum of particle: $$p_{\text{particle}} = m v = (9.1 \times 10^{-31})(10^{6}) = 9.1 \times 10^{-25}\,\text{kg m s}^{-1} \;-(1)$$
  de-Broglie wavelength of particle: $$\lambda_{\text{particle}} = \frac{h}{p_{\text{particle}}}$$

Case 2: Photon with given momentum $$p_{\text{photon}} = 10^{-27}\,\text{kg m s}^{-1}$$
  Wavelength of photon: $$\lambda_{\text{photon}} = \frac{h}{p_{\text{photon}}}$$

Required ratio of wavelengths:
$$\frac{\lambda_{\text{photon}}}{\lambda_{\text{particle}}} = \frac{\frac{h}{p_{\text{photon}}}}{\frac{h}{p_{\text{particle}}}} = \frac{p_{\text{particle}}}{p_{\text{photon}}}$$ Substituting from $$(1)$$:
$$\frac{\lambda_{\text{photon}}}{\lambda_{\text{particle}}} = \frac{9.1 \times 10^{-25}}{10^{-27}} = 9.1 \times 10^{2} = 910$$

Therefore, the wavelength of the photon is 910 times the wavelength of the particle.