Photons of wavelength 0.00150 nm undergo Compton collisions with free electrons.What is the energy of the scattered photons whose angle of scattering is 45º? A)13.3 $\times$ 10^-14 J B)9.00 $\times$ 10^-14 J C)28.0 $\times$ 10^-14 J D)18.0 $\times$ 10^-14 J E)2.14 $\times$ 10^-14 J

Question

Quizplus · Accepted Answer

We can use the formula for Compton scattering:

Δλ = h/mec(1-cosθ)

where Δλ is the change in wavelength, h is Planck's constant, me is the mass of the electron, c is the speed of light, θ is the angle of scattering.

Since we know the initial wavelength and the scattering angle, we can solve for the change in wavelength:

Δλ = 2λ sin²(θ/2) = 0.0129 nm

The energy of the scattered photon can be found using:

E = hc/λ' = hc/(λ + Δλ)

Plugging in the values, we get:

E = 9.00 x 10^-14 J

Therefore, the answer is B.

Photons of Wavelength 0 ×\times× 10-14 J B)9 ×\times× 10-14 J C)28 ×\times×

Photons of Wavelength 0 $\times$ 10^-14 J
B)9 $\times$ 10^-14 J
C)28 $\times$