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, c = 3.00 × 10⁸ m/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 possible energy loss of a photon in the Compton scattering process can be calculated using the equation:

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 angle between the incident photon and the scattered photon.

In this case, the incident photon has a wavelength of 29 pm (picometers), which is equivalent to 2.9 × 10^-11 m. The electron is stationary, so the scattered photon will be deflected at an angle of 180 degrees (i.e. backscatter).

Using these values, we can calculate the maximum possible energy loss:

ΔE = (6.626 × 10^-34 J s * 3.00 × 10^8 m/s / (2.9 × 10^-11 m)) * (1 - cos 180°)
    = 6.1 keV

Therefore, the correct answer is D (6.1 keV).

A Photon of Wavelength 29 Pm Is Scattered by a Stationary