X-rays of wavelength of 0.0656 nm undergo Compton scattering from free electrons in carbon.What is the wavelength of photons scattered at 82.5° relative to the incident beam? (h = 6.63 × 10−³⁴ J⋅s,m_e = 9.11 × 10−³¹ kg,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m) A)0.0656 nm B)0.0677 nm C)0.0848 nm D)0.0719 nm E)0.0659 nm

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Quizplus · Accepted Answer

The wavelength change 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 electron mass, $c$ is the speed of light, and $\theta$ is the scattering angle in radians. Given $h = 6.63 \times 10^{-34} \, \text{J⋅s}$, $m_e = 9.11 \times 10^{-31} \, \text{kg}$, $c = 3.00 \times 10^8 \, \text{m/s}$, and $\theta = 82.5^\circ$, we first convert the angle to radians: $\theta = 82.5^\circ \times \frac{\pi}{180} = 1.4399 \, \text{radians}$. The Compton wavelength shift constant $\lambda_C = \frac{h}{m_e c} = 2.426 \times 10^{-12} \, \text{m}$. Plugging the values into the formula gives $\Delta \lambda = 2.426 \times 10^{-12} \, \text{m} \times (1 - \cos(1.4399)) = 2.426 \times 10^{-12} \, \text{m} \times (1 - \cos(1.4399)) = 2.1 \times 10^{-12} \, \text{m} = 0.0021 \, \text{nm}$. Adding this to the initial wavelength of $0.0656 \, \text{nm}$ gives a scattered wavelength of $0.0656 \, \text{nm} + 0.0021 \, \text{nm} = 0.0677 \, \text{nm}$.

X-Rays of Wavelength of 0