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The debroglie wavelength of a particle moving with a velocity 2.25 ×10 raised to8 is eqal to that of a photon. The ratio of theri kinetic energies is.?
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De Broglie Wavelength Comparison
The de Broglie wavelength is given by the formula λ = h / p, where λ is the wavelength, h is the Planck constant, and p is the momentum of the particle.

Given Information
- Velocity of the particle = 2.25 × 10^8 m/s
- Ratio of de Broglie wavelength to that of a photon = 1

Calculating the de Broglie Wavelength
- The momentum of a particle is given by p = mv, where m is the mass of the particle and v is its velocity.
- Let's assume the mass of the particle is m.
- Therefore, the de Broglie wavelength of the particle is λ = h / (m * v).

Comparing with Photon
- The de Broglie wavelength of a photon is given by λ = c / f, where c is the speed of light and f is the frequency of the photon.
- Since the ratio of their wavelengths is given as 1, we have:
h / (m * v) = c / f
f = (m * v * c) / h

Calculating the Ratio of Kinetic Energies
- The kinetic energy of a particle is given by KE = 0.5 * m * v^2.
- The kinetic energy of a photon is given by KE = hf, where h is the Planck constant.
- Therefore, the ratio of their kinetic energies is:
KE_particle / KE_photon = (0.5 * m * v^2) / (h * f)
= (0.5 * m * v^2) / (m * v * c)
= 0.5 * v / c

Conclusion
The ratio of the kinetic energies of the particle and the photon moving with the given velocity is 0.5 * v / c.
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The debroglie wavelength of a particle moving with a velocity 2.25 ×10 raised to8 is eqal to that of a photon. The ratio of theri kinetic energies is.?
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