Nuclei in a radium-226 radioactive source emit photons whose...

Nuclei in a radium-226 radioactive source emit photons whose energy is $186 \mathrm{keV} .$ These photons are scattered by the electrons in a metal target; a detector measures the energy of the scattered photons as a function of the angle $\theta$ through which they are scattered. Find the energy of the $\gamma$ -rays scattered through $\theta=90.0^{\circ}$ and $180.0^{\circ} .$

Key Concepts

Conservation of Energy and Momentum

In Compton scattering, both energy and momentum are conserved. These conservation principles are essential for deriving the mathematical relationships of the scattering process, linking the initial and scattered photon energies to the scattering angle and the electron’s properties.

Scattering Angle

The scattering angle is the angle at which a photon is deflected as a result of its interaction with an electron. It is a crucial parameter in the Compton formula, which shows that larger angles correspond to greater increases in the photon’s wavelength and consequent energy loss.

Compton Wavelength

The Compton wavelength, defined as h/(m?c), provides a scale for the change in wavelength observed in Compton scattering. It appears in the Compton formula and represents the intrinsic length scale associated with the electron.

Compton Scattering

Compton scattering describes the inelastic interaction between a photon and a loosely bound or free electron, in which the photon transfers part of its energy to the electron and emerges with a longer wavelength. The effect demonstrates the particle-like properties of light and is governed by the conservation of energy and momentum.

Photon Energy

Photon energy is related to its frequency or wavelength by the relationship E = h? or E = hc/?. In scattering events like Compton scattering, a photon’s energy changes as it interacts with electrons, leading to a measurable shift that depends on the scattering angle.

Transcript

00:01 So in this exercise, we have a radioactive element that releases photons that have an energy of 186 kilo -electron volts.

00:12 And these photons are scattered by electrons.

00:16 And we have to find what are the energy of these scattered photons at angles 90 degrees and 180 degrees.

00:27 So the first thing we're going to have to do here is to calculate the wavelength, lambda of the incident electron this one the the electron that is emitted by the radioactive source this is going to be hc over e and this is because the energy of the the electron is given by hc over lambda according to planks formula so this is going to be 100 1 ,000 to 140 electron volts nanometers that's hc over by 187 times and 86 times 10 to the 3 electron volts.

01:13 And this is 6 .67 times 10 to the minus 3 nanometers.

01:24 So this is the wavelength of the incident photon.

01:28 And according to compton scattering, we have that the wavelength of the scattered photon, is given by the wavelength of the incident photon plus h over mec1 minus cosine of theta.

01:49 Now, for theta equals 90, we're going to have a lambda s equals lambda.

02:02 Lambda we have already calculated.

02:04 It's 6 .67 times 10 to the minus 3 nanometers plus.

02:13 H over mec, that's 2 .43 times 10 to the minus 3.

02:20 And since the cosine of 90 is 0, we're going to just multiply it by 1, so that lambda s equals 9 .1 times 10 to the minus 3 nanometers...

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