If the scattering angle of the light in Compton's scattering experiment is $90 ^ { \circ }$, what is the change in the wavelength of the light? ( $m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s }$ )
A) $3.65 \times 10 ^ { - 12 } \mathrm {~m}$
B) $1.22 \times 10 ^ { - 12 } \mathrm {~m}$
C) $4.85 \times 10 ^ { - 12 } \mathrm {~m}$
D) $2.42 \times 10 ^ { - 12 } \mathrm {~m}$

Question

Quizplus · Accepted Answer

The change in wavelength ($\Delta \lambda$) in Compton scattering is given by the formula $\Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)$, where $h$ is Planck's constant, $m_e$ is the mass of the electron, $c$ is the speed of light, and $\theta$ is the scattering angle. For a scattering angle of $90^\circ$, $\cos(90^\circ) = 0$, so the formula simplifies to $\Delta \lambda = \frac{h}{m_e c}$. Substituting the given values, $\Delta \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 3.00 \times 10^8} = 2.42 \times 10^{-12} \, \text{m}$.

If the Scattering Angle of the Light in Compton's Scattering 90∘90 ^ { \circ }90∘

If the Scattering Angle of the Light in Compton's Scattering $90 ^ { \circ }$