A 20 keV electron is brought to rest by colliding twice with target nuclei as in Fig. 40-14. (Assume the nuclei remain stationary.) The wavelength associated with the photon emitted in the second collision is 130 pm greater than that associated with the photon emitted in the first collision. (a) What is the kinetic energy of the electron after the first collision? What are (b) the wavelength ${\mathbf{\lambda }}_{\mathbf{1}}$and (c) the energy ${\mathbf{E}}_{\mathbf{1}}$associated with the first photon? What are (d) ${\mathbf{\lambda }}_{\mathbf{2}}$and (e) ${\mathbf{E}}_{\mathbf{2}}$associated with the second photon?

An electron with energy ${\mathrm{E}}_0=20\mathrm{keV}$is brought to rest by colliding twice with the target nuclei.

The wavelength associated with the second photon is $∆\mathrm{\lambda }=130\mathrm{pm}$greater than that of the first photon.

Consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{15}\times 6.63\times {10}^{-34}\right)\mathrm{keV}.\mathrm{s} \\ =41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =\left(3\times {10}^8\times {10}^{12}\right)\mathrm{pm}/\mathrm{s} \\ =3\times {10}^{20}\mathrm{pm}/\mathrm{s} \end{gathered}$$

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Using Planck's relation and the concept of conservation of energy, we can get the value of the wavelength of the initial state of the electron. Now, the equation to get the wavelength associated with the first photon after the first collision is calculated for the required wavelength value. Using this value, we can get the energy associated with the photon and also the wavelength of the second photon. Thus, using the wavelength value, the energy associated with the second photon can be calculated.

Formulae:

The energy of the photon due to Planck’s relation,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (1)

Here, $∆E$is the energy of the photon, h is the Plank’s constant, and c is the speed of light.

The roots of a quadratic equation is

$a{x}^2+bx+c=0$,

Here,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ….. (2)

Let the wavelength of the two photons be ${\lambda }_1$and ${\lambda }_2$. Here, the wavelength ${\lambda }_2$is,

${\lambda }_2={\lambda }_1+\Delta \lambda$

Now, according to the energy conservation and equation (1), the energy before and after a collision can be given as follows:

$$\begin{gathered} \frac{hc}{{\lambda }_0}=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_1+∆\lambda } \\ \frac{1}{{\lambda }_0}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_1+∆\lambda } \\ \frac{1}{{\lambda }_0}=\frac{{\lambda }_1+∆\lambda +{\lambda }_1}{\left({\lambda }_1+∆\lambda \right)\left({\lambda }_1\right)} \end{gathered}$$

$$\begin{gathered} {\lambda }_1^2+{\lambda }_1\Delta \lambda =2{\lambda }_0{\lambda }_1+{\lambda }_0\Delta \lambda \\ {\lambda }_1^2+{\lambda }_1(\Delta \lambda -2{\lambda }_0)-{\lambda }_0\Delta \lambda =0 \end{gathered}$$

The above equation is a quadratic equation, whose roots can be given using equation (2) as:

For $a=1,b=(\Delta \lambda -2{\lambda }_0),c=-{\lambda }_0\Delta \lambda$

$$\begin{gathered} {\mathrm{\lambda }}_1=\frac{-\left(∆\mathrm{\lambda }-2{\mathrm{\lambda }}_0\right)\pm \sqrt{{\left(∆\mathrm{\lambda }-2{\mathrm{\lambda }}_0\right)}^2+4\left({\mathrm{\lambda }}_0∆\mathrm{\lambda }\right)}}{2} \\ =\frac{-\left(∆\mathrm{\lambda }-2{\mathrm{\lambda }}_0\right)\pm \sqrt{{\left(∆\mathrm{\lambda }\right)}^2+{\left(2{\mathrm{\lambda }}_0\right)}^2+4\left({\mathrm{\lambda }}_0∆\mathrm{\lambda }\right)+4\left({\mathrm{\lambda }}_0∆\mathrm{\lambda }\right)}}{2} \\ {\mathrm{\lambda }}_1=\frac{-\left(∆\mathrm{\lambda }/{\mathrm{\lambda }}_0-2\right)\pm \sqrt{{\left(∆\mathrm{\lambda }/{\mathrm{\lambda }}_0\right)}^2+4}}{2/{\mathrm{\lambda }}_0} \end{gathered}$$ ..........

(3)

But, using the given data in equation (1), the wavelength associated with energy before collision can be calculated as follows:

$$\begin{gathered} {\mathrm{\lambda }}_0=\frac{\mathrm{hv}}{∆\mathrm{E}} \\ =\frac{\left(41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3\times {10}^{20}\mathrm{pm}/\mathrm{s}\right)}{20\mathrm{keV}} \\ =\frac{1240\mathrm{keV}.\mathrm{pm}}{20\mathrm{keV}} \\ =62\mathrm{pm} \end{gathered}$$

Now, using this above value in equation (3) for the positive condition of equation (3)$\left(\because {\lambda }_1>0\right)$, the value of the wavelength associated with the first photon is as follow.

$$\begin{gathered} {\mathrm{\lambda }}_1=\frac{-\left({\displaystyle \raisebox{1ex}{$130\mathrm{pm}$}\!\left/ \!\raisebox{-1ex}{$62{\mathrm{pm}}^2$}\right.}\right)+\sqrt{{\left({\displaystyle \raisebox{1ex}{$130\mathrm{pm}$}\!\left/ \!\raisebox{-1ex}{$62\mathrm{pm}$}\right.}\right)}^2+4}}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$62\mathrm{pm}$}\right.}} \\ =\frac{-\left(2.0967-2\right)+\sqrt{4.39462+4}}{2/62\mathrm{pm}} \\ =\frac{-\left(0.096774\right)+\sqrt{8.39462}}{2/62\mathrm{pm}} \end{gathered}$$

$$\begin{gathered} {\mathrm{\lambda }}_2=\left[-\left(0.096774\right)+\left(2.897665\right)\right]\frac{62\mathrm{pm}}{2} \\ =\left(2.800891\right)\times 31 \\ =86.82\mathrm{pm} \\ \approx 87\mathrm{pm} \end{gathered}$$

Thus, the energy of the electron after its first deceleration is given by:

$$\begin{gathered} \mathrm{K}={\mathrm{K}}_0-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1} \\ =20\mathrm{keV}-\frac{1240\mathrm{keV}.\mathrm{pm}}{87\mathrm{pm}} \\ =5.7\mathrm{keV} \end{gathered}$$

Hence, the value of the required kinetic energy after its first collision is 5.7keV .

From the calculations of part (a), the wavelength associated with the first photon for the first collision is found to be 87 pm.

Using the wavelength value of part (b) in equation (1), the energy $E_1$associated with the first photon after the first collision as follows:

$$\begin{gathered} {\mathrm{E}}_1=\frac{1240\mathrm{keV}.\mathrm{pm}}{87\mathrm{pm}} \\ =14\mathrm{keV} \end{gathered}$$

Hence, the value of the energy is 14. keV.

Now, using the given condition, the wavelength associated with the second photon can be calculated as follows:

$$\begin{gathered} {\mathrm{\lambda }}_2=\left(87+130\right)\mathrm{pm} \\ =2.2\times {10}^2\mathrm{pm} \end{gathered}$$

Hence, the value of the wavelength is $2.2\times {10}^2\mathrm{pm}$.

Using the wavelength value of part (b) in equation (1), the energy $E_2$associated with the second photon after the second collision is as follow.

$$\begin{gathered} {\mathrm{E}}_2=\frac{1240\mathrm{keV}.\mathrm{pm}}{2.2\times {10}^2\mathrm{pm}} \\ =5.7\mathrm{keV} \end{gathered}$$

Hence, the value of the energy is 5.7 keV.