Chapter 26, Problem 55QAP

To determine

(a)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is ${0.00}^0$.

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.140 nm and the kinetic energy of the scattered electron is 0 if the scattering angle is ${0.00}^0$.

Explanation of Solution

Given:

Initial wavelength ${\lambda }_i=0.140\text{ nm}$ and the scattering angle is ${0.00}^0$

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=\frac{h}{m_{e}c}(1-\cos \theta )$ (1.1)

Now the energy of the scattered photon is

  $E_f=\frac{hc}{{\lambda }_f}$ (1.2)

Kinetic energy of the scattered electron is given by

  $K=E_i-E_f=hc\left(\frac{{\lambda }_f-{\lambda }_i}{{\lambda }_f{\lambda }_i}\right)=hc\left(\frac{\Delta {\lambda }_{}}{{\lambda }_f{\lambda }_i}\right)$(1.3)

Calculation:

For $\theta ={0.00}^0$ using Eq. (1.1) we get,

  $\begin{array}{c}\Delta \lambda =\frac{h}{m_{e}c}(1-\cos 0°)\\ =\frac{h}{m_{e}c}(1-1)\\ =0\end{array}$

So,

  ${\lambda }_f={\lambda }_i=0.140\text{ nm}$

And using the value of $\Delta \lambda =0$ in Eq. (1.3), we get

K = 0.

Conclusion:

So the wavelength of the Compton scattered photon is 0.140 nm and the kinetic energy of the scattered electron is 0

To determine

(b)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is ${30.00}^0$

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.140325 nm and the kinetic energy of the scattered electron is 20.5 eV

Explanation of Solution

Given:

Initial wavelength ${\lambda }_i=0.140\text{ nm}$ and the scattering angle is ${30.00}^0$

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=\frac{h}{m_{e}c}(1-\cos \theta )$(1.1)

Now the energy of the scattered photon is

  $E_f=\frac{hc}{{\lambda }_f}$ (1.2)

Kinetic energy of the scattered electron is given by

  $K=E_i-E_f=hc\left(\frac{{\lambda }_f-{\lambda }_i}{{\lambda }_f{\lambda }_i}\right)=hc\left(\frac{\Delta {\lambda }_{}}{{\lambda }_f{\lambda }_i}\right)$(1.3)

Calculation:

For $\theta ={30}^0$ using Eq. (1.1) we get

  $\begin{array}{l}\Delta \lambda =({\lambda }_f-{\lambda }_i)=2.43\times {10}^{-12}(1-\cos {30}^0)\text{ m}=0.325\times {10}^{-3}\text{ nm}\\ {\lambda }_f=0.140+0.325\times {10}^{-3}\text{ nm = 0}\text{.140325 nm}\end{array}$

Now using the values of $\Delta \lambda$, ${\lambda }_f$ and ${\lambda }_i$ in Eq. (1.3) we get

  $\begin{array}{l}K=hc\left(\frac{\Delta \lambda }{{\lambda }_f{\lambda }_i}\right)=\frac{6.63\times {10}^{-34}\times 3\times {10}^8\times 0.325\times {10}^{-12}}{0.140325\times 0.140\times {10}^{-18}}\text{ J}\\ \text{or, K = }\frac{329\times {10}^{-20}}{1.6\times {10}^{-19}}\text{eV}=20.5\text{ eV}\end{array}$

[Above we have used 1 nm = 10-9 m and $1\text{eV}=1.6\times {10}^{-19}\text{J}$ ]

Conclusion:

So the wavelength of the Compton scattered photon is 0.140325 nm and the kinetic energy of the scattered electron is 20.5 eV.

To determine

(c)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is ${45.00}^0$.

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.140712 nm and the kinetic energy of the scattered electron is 44.9 eV

Explanation of Solution

Given:

Initial wavelength ${\lambda }_i=0.140\text{ nm}$ and the scattering angle is ${45.00}^0$

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=\frac{h}{m_{e}c}(1-\cos \theta )$(1.1)

Now the energy of the scattered photon is

  $E_f=\frac{hc}{{\lambda }_f}$ (1.2)

Kinetic energy of the scattered electron is given by,

  $K=E_i-E_f=hc\left(\frac{{\lambda }_f-{\lambda }_i}{{\lambda }_f{\lambda }_i}\right)=hc\left(\frac{\Delta {\lambda }_{}}{{\lambda }_f{\lambda }_i}\right)$(1.3)

Calculation:

For $\theta ={45.00}^0$ using Eq. (1.1)

  $\begin{array}{l}\Delta \lambda =({\lambda }_f-{\lambda }_i)=2.43\times {10}^{-12}(1-\cos {45}^0)\text{ m}\\ \text{or, }{\lambda }_f=0.140712\text{ nm}\end{array}$

Now using the values of $\Delta \lambda$, ${\lambda }_f$ and ${\lambda }_i$ in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and $1\text{eV}=1.6\times {10}^{-19}\text{J}$ ] we get

  $K=44.9\text{ eV}$

Conclusion:

Wavelength of the Compton scattered photon is 0.140712 nm and the kinetic energy of the scattered electron is 44.9 eV

To determine

(d)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is ${60.00}^0$

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.141215 nm and the kinetic energy of the scattered electron is 75.32 eV if the scattering angle is ${60.00}^0$.

Explanation of Solution

Given:

Initial wavelength ${\lambda }_i=0.140\text{ nm}$ and the scattering angle is ${60.00}^0$

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=\frac{h}{m_{e}c}(1-\cos \theta )$(1.1)

Now the energy of the scattered photon is

  $E_f=\frac{hc}{{\lambda }_f}$ (1.2)

Kinetic energy of the scattered electron is given by,

  $K=E_i-E_f=hc\left(\frac{{\lambda }_f-{\lambda }_i}{{\lambda }_f{\lambda }_i}\right)=hc\left(\frac{\Delta {\lambda }_{}}{{\lambda }_f{\lambda }_i}\right)$(1.3)

Calculation:

For $\theta ={60.00}^0$ using Eq. (1.1)

  $\begin{array}{l}\Delta \lambda =({\lambda }_f-{\lambda }_i)=2.43\times {10}^{-12}(1-\cos {60}^0)\text{ m = 0}\text{.001215 nm}\\ \text{or, }{\lambda }_f=0.141215\text{ nm}\end{array}$

Now using the values of $\Delta \lambda$, ${\lambda }_f$ and ${\lambda }_i$ in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and $1eV=1.6\times {10}^{-19}\text{J}$ ] we get,

  $K=75.32\text{ eV}$

Conclusion:

Wavelength of the Compton scattered photon is 0.141215 nm and the kinetic energy of the scattered electron is 75.32 eV

To determine

(e)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is ${90.00}^0$

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.14243 nm and the kinetic energy of the scattered electron is 151.32 eV

Explanation of Solution

Given:

Initial wavelength ${\lambda }_i=0.140\text{ nm}$ and the scattering angle is 90.00⁰

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=\frac{h}{m_{e}c}(1-\cos \theta )$ (1.1)

Now the energy of the scattered photon is

  $E_f=\frac{hc}{{\lambda }_f}$ (1.2)

Kinetic energy of the scattered electron is given by,

  $K=E_i-E_f=hc\left(\frac{{\lambda }_f-{\lambda }_i}{{\lambda }_f{\lambda }_i}\right)=hc\left(\frac{\Delta {\lambda }_{}}{{\lambda }_f{\lambda }_i}\right)$(1.3)

Calculation:

For θ = 90⁰ using Eq. (1.1)

  $\begin{array}{l}\Delta \lambda =({\lambda }_f-{\lambda }_i)=2.43\times {10}^{-12}(1-\cos {90}^0)\text{ m = }2.43\times {10}^{-12}\text{ m}\\ \text{or, }{\lambda }_f=0.14243\text{ nm}\end{array}$

Now using the values of $\Delta \lambda$, ${\lambda }_f$ and ${\lambda }_i$ in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and $1\text{eV}=1.6\times {10}^{-19}\text{J}$ ] we get

  $K=151.32\text{ eV}$

Conclusion:

Wavelength of the Compton scattered photon is 0.14243 nm and the kinetic energy of the scattered electron is 151.32 eV.

To determine

(f)

Wavelength of the Compton scattered photon and the kinetic energy of the scattered electron if the scattering angle is ${180.00}^0$.

Answer to Problem 55QAP

Wavelength of the Compton scattered photon is 0.14486 nm and the kinetic energy of the scattered electron is 297.43 eV

Explanation of Solution

Given:

Initial wavelength ${\lambda }_i=0.140\text{ nm}$ and the scattering angle is 180.00⁰

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by

  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=\frac{h}{m_{e}c}(1-\cos \theta )$(1.1)

Now the energy of the scattered photon is

  $E_f=\frac{hc}{{\lambda }_f}$ (1.2)

Kinetic energy of the scattered electron is given by,

  $K=E_i-E_f=hc\left(\frac{{\lambda }_f-{\lambda }_i}{{\lambda }_f{\lambda }_i}\right)=hc\left(\frac{\Delta {\lambda }_{}}{{\lambda }_f{\lambda }_i}\right)$(1.3)

Calculation:

For θ = 180⁰ using Eq. (1.1)

  $\begin{array}{l}\Delta \lambda =({\lambda }_f-{\lambda }_i)=2.43\times {10}^{-12}(1-\cos {180}^0)\text{ m = }4.86\times {10}^{-12}\text{ m}\\ \text{or, }{\lambda }_f=0.14486\text{ nm}\end{array}$

Now using the values of $\Delta \lambda$, ${\lambda }_f$ and ${\lambda }_i$ in Eq. (1.3) and using the conversions as used earlier in part (b) above [1 nm = 10-9 m and $1eV=1.6\times {10}^{-19}\text{J}$ ] we get

  $K=\text{ 297}\text{.43 eV}$

Conclusion:

Wavelength of the Compton scattered photon is 0.14486 nm and the kinetic energy of the scattered electron is 297.43 eV