A certain photon,after being scattered from a free electron that was at rest,moves at an angle of 120° with respect to the incident direction.If the wavelength of the incident photon is 0.591 nm, what is the wavelength of the scattered photon? $\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.$ $c = 3.00 \times 10^8 \mathrm {~m} / \mathrm { s }$ $h = 6.626 \times 10^{ - 34} \mathrm {~J} \cdot \mathrm { s } )$
A)0.380 nm
B)0 nm
C)0.550 nm
D)0.591 nm
E)0.595 nm

Question

Quizplus · Accepted Answer

The wavelength of the scattered photon can be calculated using the Compton scattering formula: $\lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta)$, where $\lambda'$ is the wavelength of the scattered photon, $\lambda$ is the initial wavelength (0.591 nm), $h$ is Planck's constant, $m_e$ is the mass of the electron, $c$ is the speed of light, and $\theta$ is the scattering angle (120°). Substituting the given values and converting $\lambda$ to meters, the calculation yields a wavelength for the scattered photon slightly longer than the initial wavelength, which corresponds to choice E, 0.595 nm.

A Certain Photon,after Being Scattered from a Free Electron That (melectron =9.11×10−31 kg,\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.(melectron ​=9.11×10−31 kg,

A Certain Photon,after Being Scattered from a Free Electron That $\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , \right.$