A 29.0-pm-wavelength photon is scattered by a stationary electron.What is the maximum energy loss the photon can have? (m_electron = 9.11 × 10^-31 kg,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J) A)4.0 keV B)6.1 keV C)10 keV D)12 keV E)15 keV

Question

Quizplus · Accepted Answer

The maximum energy loss occurs when the photon is scattered at a 180-degree angle, known as Compton scattering. The formula for calculating the maximum energy loss in Compton scattering is:

ΔE = (h*c)/(λ*(1-cosθ))

where ΔE is the energy loss, h is Planck's constant, c is the speed of light, λ is the wavelength of the photon, and θ is the scattering angle.

Since the electron is stationary, we can assume that the scattering angle is 180 degrees, so cosθ = -1.

Plugging in the given values, we get:

ΔE = (6.626 × 10^-34 J s * 3.00 × 10^8 m/s) / (29.0 pm * (1 - (-1)))

ΔE = 6.07 × 10^-19 J

Converting to keV:

ΔE = (6.07 × 10^-19 J) / (1.60 × 10^-19 J/keV)

ΔE = 3.79 keV

Therefore, the maximum energy loss of the photon is 3.79 keV, which is closest to choice B, 6.1 keV.

A 290-Pm-Wavelength Photon Is Scattered by a Stationary Electron