Mention the change in wave length of the photon after it collides with free electron?? Is the rule of particle can be applied here?

Jadon Johnson
2022-11-18
Answered

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kavdawg8w8

Answered 2022-11-19
Author has **20** answers

You are probably familiar with the relation

$c=\lambda f$

where c is the speed of light, $\lambda $ the wavelength and f the frequency of a photon.

The general formula for Compton scattering is normally given as

${\lambda}^{\prime}={\lambda}_{C}(1-\mathrm{cos}\alpha )+\lambda $

where ${\lambda}^{\prime}$ is the wavelength of the scattered photon, $\lambda $ the initial wavelength, ${\lambda}_{C}$ the Compton wavelength ${\lambda}_{C}=\frac{h}{{m}_{e}c}$ and $\alpha $ is the angle at which the photon scatters.

From this formula, we can see that the wavelength of a photon increases when it scatters at an electron. To visualize, the photon gives some of its energy to the electron. The energy of a photon is given as

$E=hf$

or $E=\frac{hc}{\lambda}$ so if the wavelength increases, the energy of the photon decreases. To illustrate further, if you shoot blue light at a stationary electron, you end up with scattered red light.

For the case of electromagetic radiation (light!), you can convert between frequency and wavelength using $c=\lambda f$. Those formulas combined allow you to find the frequency and wavelength of a scattered photon.

$c=\lambda f$

where c is the speed of light, $\lambda $ the wavelength and f the frequency of a photon.

The general formula for Compton scattering is normally given as

${\lambda}^{\prime}={\lambda}_{C}(1-\mathrm{cos}\alpha )+\lambda $

where ${\lambda}^{\prime}$ is the wavelength of the scattered photon, $\lambda $ the initial wavelength, ${\lambda}_{C}$ the Compton wavelength ${\lambda}_{C}=\frac{h}{{m}_{e}c}$ and $\alpha $ is the angle at which the photon scatters.

From this formula, we can see that the wavelength of a photon increases when it scatters at an electron. To visualize, the photon gives some of its energy to the electron. The energy of a photon is given as

$E=hf$

or $E=\frac{hc}{\lambda}$ so if the wavelength increases, the energy of the photon decreases. To illustrate further, if you shoot blue light at a stationary electron, you end up with scattered red light.

For the case of electromagetic radiation (light!), you can convert between frequency and wavelength using $c=\lambda f$. Those formulas combined allow you to find the frequency and wavelength of a scattered photon.

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If an x-ray source has an activity of ${10}^{9}$ becquerel then by the relation:

$Number\text{}of\text{}photons=Activity\ast yield$

is the Number of photons equal to ${10}^{9}$ ?

$Number\text{}of\text{}photons=Activity\ast yield$

is the Number of photons equal to ${10}^{9}$ ?

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