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 60º? A) 13.3 × 10^-14 J B) 3.27 × 10^-14 J C) 28.0 × 10^-14J D) 7.36 × 10^-14 J E) 2.14 × 10^-14 J

Question

Quizplus · Accepted Answer

We use the Compton scattering formula:

$\Delta\lambda = \dfrac{h}{m_{e}c}(1-\cos	heta)$

Where $\Delta\lambda$ is the change in wavelength, $h$ is Planck's constant, $m_{e}$ is the mass of an electron, $c$ is the speed of light, and $	heta$ is the angle of scattering.

The change in wavelength is given by:

$\Delta\lambda = \lambda_{f} - \lambda_{i}$

Where $\lambda_{f}$ is the final wavelength (i.e. scattered wavelength) and $\lambda_{i}$ is the initial wavelength.

Using the given initial wavelength, we can calculate the initial energy of the photon:

$E_{i} = hc/\lambda_{i}$

Once we know the scattered wavelength, we can calculate the final energy of the photon using:

$E_{f} = hc/\lambda_{f}$

Since energy is conserved, we can equate $E_{i}$ and $E_{f}$:

$hc/\lambda_{i} = hc/\lambda_{f}$

Solving for $\lambda_{f}$, we get:

$\lambda_{f} = \lambda_{i}/(1-\cos	heta)$

Plugging in the given values, we get:

$\lambda_{f} = 0.00150/(1-\cos60º) = 0.00404$ nm

Now that we know the final wavelength, we can calculate the energy of the photon after scattering:

$E_{f} = hc/\lambda_{f} = 3.08 	imes 10^{-16}$ J

Therefore, the correct answer is D) 7.36 × 10^−14 J (rounded to two significant figures).

Photons of Wavelength 0