A photon of wavelength 29 pm is scattered by a stationary electron. What is the maximum possible energy loss of the photon? (mel = 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 the photon occurs when it is scattered at a 180-degree angle, which gives the maximum momentum transfer to the electron. In this case, we can use the Compton scattering formula:

λ' - λ = h/mc * (1 - cosθ)

where λ is the initial wavelength of the photon, λ' is the scattered wavelength, h is Planck's constant, m is the mass of the electron, c is the speed of light, and θ is the scattering angle.

Since the electron is initially at rest, we can assume that the scattering angle is 180 degrees. Plugging in the given values, we get:

λ' - λ = (6.626 × 10^-34 J ∙ s)/(9.11 × 10^-31 kg) * (1 - cos180°)
λ' - λ = 0.00731 pm

Therefore, the maximum possible energy loss of the photon is:

ΔE = hc/λ - hc/λ'
ΔE = (6.626 × 10^-34 J ∙ s * 3.00 × 10^8 m/s)/(29 × 10^-12 m) - (6.626 × 10^-34 J ∙ s * 3.00 × 10^8 m/s)/(0.00731 × 10^-12 m)
ΔE = 6.12 keV

The closest choice is (D) 6.1 keV.

A Photon of Wavelength 29 Pm Is Scattered by a Stationary