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Step-by-Step Solution
Step 1: Write down the known quantities
• Mass of the particle, m = 9.1 \times 10^{-31}\,\text{kg}
• Speed of the particle, v = 10^{6}\,\text{m/s}
• Linear momentum of the photon, P_{\text{photon}} = 10^{-27}\,\text{kg m/s}
• Planck’s constant, h = 6.6 \times 10^{-34}\,\text{J s}
Step 2: Write the expression for the wavelength of the photon
A photon’s wavelength \lambda_{1} can be calculated using the relation between momentum ( P ) and Planck’s constant ( h ):
\lambda_{1} = \frac{h}{P_{\text{photon}}}.
Step 3: Substitute values for the photon's wavelength
Substitute h = 6.6 \times 10^{-34}\,\text{J s} and P_{\text{photon}} = 10^{-27}\,\text{kg m/s} :
\lambda_{1} = \frac{6.6 \times 10^{-34}}{10^{-27}} = 6.6 \times 10^{-7}\,\text{m}.
Step 4: Write the expression for the particle's wavelength
The de Broglie wavelength \lambda_{2} of a particle of mass m moving with speed v is given by:
\lambda_{2} = \frac{h}{mv}.
Step 5: Substitute values for the particle's wavelength
Now, substitute h = 6.6 \times 10^{-34}\,\text{J s} , m = 9.1 \times 10^{-31}\,\text{kg} , and v = 10^{6}\,\text{m/s} :
\lambda_{2} = \frac{6.6 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (10^{6})}.
Combine the terms in the denominator:
(9.1 \times 10^{-31}) \times (10^{6}) = 9.1 \times 10^{-25}.
Hence,
\lambda_{2} = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-25}} \approx 7.25 \times 10^{-10}\,\text{m}.
Step 6: Determine the ratio of the photon’s wavelength to the particle’s wavelength
\frac{\lambda_{1}}{\lambda_{2}}
= \frac{6.6 \times 10^{-7}}{7.25 \times 10^{-10}}
\approx 910.
So, the wavelength of the photon is 910 times the wavelength of the particle.
Final Answer:
The wavelength of the photon is 910 times the wavelength of the particle.
