A certain photon after being scattered from a free electron at rest moves at an angle of 120° with respect to the incident direction. If the wavelength of the incident photon is 0.611 nm, what is the kinetic energy of the recoiling electron? A) 1.5 × 10^-18 J B) 2.1 × 10^-18 J C) 3.1 × 10^-18 J D) 4.3 × 10^-18 J E) 5.2 × 10^-18 J

Question

Quizplus · Accepted Answer

The scattering of a photon by a free electron can be modeled as an elastic collision between the photon and the electron. In an elastic collision, energy is conserved, so the total energy of the system before the collision is equal to the total energy after the collision. The energy of a photon can be related to its wavelength by the equation:

E = hc/λ

where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

In this problem, we are given the wavelength of the incident photon (λ = 0.611 nm) and the angle by which it is scattered (θ = 120°). From the law of conservation of momentum, we know that the momentum of the recoiling electron must be equal and opposite to the momentum of the scattered photon. We can use this fact, along with the conservation of energy, to solve for the kinetic energy of the electron.

The momentum of a photon is given by:

p = h/λ

where p is the momentum of the photon. The momentum of the scattered photon can be related to the momentum of the recoiling electron by the equation:

p = mev

where me is the mass of the electron, v is its velocity, and p and v are in the same direction.

By conservation of momentum, we have:

p = 2p sin(θ/2)

where θ is the scattering angle. Substituting in the expressions for p and mev above, we can solve for v:

v = p/me = (2h/λme) sin(θ/2)

Once we have the velocity of the recoiling electron, we can use the equation for kinetic energy:

KE = (1/2)mev^2

to solve for its kinetic energy.

Plugging in the given values, we get:

KE = (1/2)(9.109 × 10^-31 kg)(3.00 × 10^8 m/s)(2π/h)(0.611 × 10^-9 m)^2 sin^2(60°)

KE = 2.1 × 10^-18 J

Therefore, the answer is B.

A Certain Photon After Being Scattered from a Free Electron