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A monochromatic x-ray beam whose wavelength is 55.8 pm is scattered through 46 deg Find the wavelength of the scattered beam

Question

A monochromatic x-ray beam whose wavelength is 55.8 pm is scattered through 46 deg. Find the wavelength of the scattered beam.

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Solution

To solve this problem, we can use the Compton scattering formula, which is given by:

Δλ = λ' - λ = h/(m_e * c) * (1 - cos θ)

where:

  • Δλ is the change in wavelength,
  • λ' is the wavelength of the scattered beam,
  • λ is the initial wavelength,
  • h is Planck's constant (6.62607015 × 10^-34 m^2 kg / s),
  • m_e is the electron rest mass (9.10938356 × 10^-31 kilograms),
  • c is the speed of light (299792458 m/s),
  • θ is the scattering angle.

We are given:

  • λ = 55.8 pm = 55.8 * 10^-12 m,
  • θ = 46 degrees = 46 * π/180 radians = 0.80285 radians.

We need to find λ', the wavelength of the scattered beam.

First, we calculate Δλ using the Compton scattering formula:

Δλ = h/(m_e * c) * (1 - cos θ) = (6.62607015 × 10^-34 m^2 kg / s) / (9.10938356 × 10^-31 kilograms * 299792458 m/s) * (1 - cos(0.80285 radians)) = 2.4263102389 * 10^-12 m * (1 - 0.6845471059286886) = 2.4263102389 * 10^-12 m * 0.3154528940713114 = 7.6539828008 * 10^-13 m = 0.76539828008 pm.

Then, we find λ' by adding Δλ to the initial wavelength λ:

λ' = λ + Δλ = 55.8 pm + 0.76539828008 pm = 56.56539828008 pm.

So, the wavelength of the scattered beam is approximately 56.57 pm.

This problem has been solved

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