Question

X-rays with an energy of 265 keV undergo Compton scattering from a target. If the scattered rays are deflected at 41.0° relative to the direction of the incident rays, find each of the following. (a) the Compton shift at this angle _________nm (b) the energy of the scattered x-ray __________keV (c) the kinetic energy of the recoiling electron ___________keV

Answer 1

The first one:

the energy of the scattered x-ray

The answer for last on:

Kinetic energy of the recoiling electron

Answer 2

Answers:

(a) Compton shift

The Compton Shift
`\Delta \lambda` in wavelength when the photons are scattered is given by the following equation:

`\Delta \lambda=\lambda' - \lambda_(o)=\lambda_(c)(1-cos\theta)` (1)

Where:

`\lambda_(c)=2.43(10)^(-12) m` is a constant whose value is given by
`(h)/(m_(e).c)`, being
`h=4.136(10)^(-15)eV.s` the Planck constant,
`m_(e)` the mass of the electron and
`c=3(10)^(8)m/s` the speed of light in vacuum.

`\theta=41\°` the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are deflected at
`41\°`:

`\Delta \lambda=\lambda' - \lambda_(o)=\lambda_(c)(1-cos(41\°))` (2)

`\Delta \lambda=\lambda' - \lambda_(o)=5.950(10)^(-13)m` (3)

But we are asked to express this in
`nm`, so:

`\Delta \lambda=5.950(10)^(-13)m.(1nm)/((10)^(-9)m)`

`\Delta \lambda=0.000595nm` (4)

(b) the energy of the scattered x-ray

The initial energy
`E_(o)=265keV=265(10)^(3)eV` of the photon is given by:

`E_(o)=(h.c)/(\lambda_(o))` (5)

From this equation (5) we can find the value of
`\lambda_(o)`:

`\lambda_(o)=(h.c)/(E_(o))` (6)

`\lambda_(o)=((4.136(10)^(-15)eV.s)(3(10)^(8)m/s))/(265(10)^(3)eV)`

`\lambda_(o)=4.682(10)^(-12)m` (7)

Knowing the value of
`\Delta \lambda` and
`\lambda_(o)`, let's find
`\lambda'`:

`\Delta \lambda=\lambda' - \lambda_(o)`

Then:

`\lambda'=\Delta \lambda+\lambda_(o)` (8)

`\lambda'=5.950(10)^(-13)m+4.682(10)^(-12)m`

`\lambda'=5.277(10)^(-12)m` (9)

Knowing the wavelength of the scattered photon
`\lambda'` , we can find its energy
`E'` :

`E'=(h.c)/(\lambda')` (10)

`E'=((4.136(10)^(-15)eV.s)(3(10)^(8)m/s))/(5.277(10)^(-12)m)`

`E'=235.121keV` (11) This is the energy of the scattered photon

(c) Kinetic energy of the recoiling electron

If we want to know the kinetic energy of the recoiling electron
`E_(e)`, we have to calculate all the energy lost by the photon in the wavelength shift, which is:

`K_(e)=E_(o)-E'` (12)

`K_(e)=265keV-235.121keV`

Finally we obtain the kinetic energy of the recoiling electron:

`E_(e)=29.878keV`