A photon of wavelength 29 pm is scattered by a stationary electron.What is the maximum possible energy loss of the photon? (m_el = 9.11 × 10^-31 kg,h = 6.626 × 10^-34 J • s, A) 4.0 keV B) 7.0 keV C) 10 keV D) 6.1 keV E) 12 keV

Question

Quizplus · Accepted Answer

The maximum energy loss of a photon scattered by an electron is given by the Compton formula:

ΔE = (h c)/(λ(1 - cosθ))

where h is Planck's constant, c is the speed of light, λ is the wavelength of the incident photon, and θ is the scattering angle. Since the electron is stationary, the scattering angle is 180 degrees.

Substituting the given values, we get:

ΔE = (6.626 × 10^-34 J•s × 3 × 10^8 m/s)/(29 × 10^-12 m(1 - cos 180°))
    ≈ 6.1 keV

Therefore, the maximum possible energy loss of the photon is 6.1 keV, which is closest to choice D.

A Photon of Wavelength 29 Pm Is Scattered by a Stationary