Question:medium

X- rays of wavelength 15 pm are scattered from a target. The wavelength of the X-rays scattered through 60° is (Given compton wavelength = 2.426 pm)

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Remember that in Compton scattering, the scattered photon always has a longer wavelength (and therefore lower energy) than the incident photon, except for forward scattering (\(\theta=0\)), where the wavelength does not change. This helps you eliminate any options that suggest a decrease in wavelength.
Updated On: Feb 20, 2026
  • 13.787 pm (pm is pico metre)
  • 16.213 pm (pm is pico metre)
  • 15.5 pm (pm is pico metre)
  • 1.6213 pm (pm is pico metre)
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The Correct Option is B

Solution and Explanation

Concept: This problem pertains to Compton scattering, a phenomenon where X-rays or gamma rays interact with charged particles, typically electrons. This interaction causes a shift in the wavelength of the scattered radiation, increasing it. The magnitude of this wavelength shift is contingent upon the angle at which the radiation is scattered.
Formula: The Compton scattering equation quantifies the change in wavelength, \( \Delta\lambda \):
\[ \Delta\lambda = \lambda' - \lambda = \lambda_c (1 - \cos\theta) \]
In this equation, \( \lambda' \) represents the wavelength of the scattered radiation, \( \lambda \) is the wavelength of the incident radiation, \( \lambda_c \) is the Compton wavelength of the electron (\(h/m_e c\)), and \( \theta \) denotes the scattering angle.
Calculation:
1. Given Values:
Incident wavelength, \( \lambda = 15 \) pm.
Compton wavelength, \( \lambda_c = 2.426 \) pm.
Scattering angle, \( \theta = 60^\circ \).
2. Change in Wavelength (\( \Delta\lambda \)):
Calculate the cosine of the scattering angle: \( \cos(60^\circ) = 0.5 \).
Apply the Compton formula with the given values:
\[ \Delta\lambda = 2.426 \, \text{pm} \times (1 - 0.5) \] \[ \Delta\lambda = 2.426 \, \text{pm} \times 0.5 = 1.213 \, \text{pm} \]
3. Scattered Wavelength (\( \lambda' \)):
Determine the new wavelength by adding the change in wavelength to the original wavelength.
\[ \lambda' = \lambda + \Delta\lambda \] \[ \lambda' = 15 \, \text{pm} + 1.213 \, \text{pm} = 16.213 \, \text{pm} \]
Result: The scattered wavelength of the X-rays at a 60° angle is 16.213 pm.
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