In a Compton scattering experiment,the scattered photon has a wavelength of 6.55 pm.If the scattering angle is 50$^{\circ}$,what is the energy of the incident photon?
A) 1.43 MeV
B) 218 keV
C) 380 keV
D) 117 keV
E) 66.3 keV

Question

Quizplus · Accepted Answer

We can use the formula for Compton scattering to solve for the energy of the incident photon: 
Δλ = λ' - λ = (h/mc)(1 - cosθ)

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

Rearranging and solving for the energy of the incident photon, we get: 
E = hc/λ = (h/mc)(1/(1 - (λ'/λ)cosθ))

Plugging in the given values gives: 
E = (6.626 x 10^-34 J.s / (9.109 x 10^-31 kg x 3 x 10^8 m/s))(1/(1 - (6.55 x 10^-12 m / λ)cos50°))

Solving for λ using the wavelength formula λ = hc/E and converting it to meters gives:
λ = (6.626 x 10^-34 J.s / (E x 1.602 x 10^-19 J/eV))(10^-12 m/nm)

Substituting for λ in the above expression for the incident photon energy and solving gives E = 218 keV. Therefore, the best choice is B.

In a Compton Scattering Experiment,the Scattered Photon Has a Wavelength ∘^{\circ}∘

In a Compton Scattering Experiment,the Scattered Photon Has a Wavelength $^{\circ}$