Chapter 42, Problem 42SP

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To determine

The de Broglie wavelength of an electron if it has been accelerated through a potential difference of $1.0\text{ MV}$.

Answer to Problem 42SP

Solution:

$8.7\times {10}^{-13}\text{ m}$

Explanation of Solution

Given data:

The electron is accelerated through the potential difference of $1.0\text{ MV}$.

Formula used:

Write the expression for de Broglie wavelength:

$\lambda =\frac{h}{p}$

Here, $h$ is the Planck’s constant and $p$ is the momentum.

Write the expression for the relativistic energy:

$E=KE+m_0{c}^2$

Also, the modified form of the relativistic energy is as follows:

$E^2=m_0{}^2{c}^4+p^2{c}^2$

The expression for the energy associated with a moving charge under influence of an electric field is

$E_{\text{electric field}}=qV$

Here, $m_0$ is the rest mass of the object, $c$ is the speed of the light, $e$ is the charge of the electron, and $V$ is the potential difference.

Explanation:

Recall the expression for the energy associated with a moving charge under influence of an electric field is

$E_{\text{electric field}}=qV$

Recall the expression for the relativistic energy:

$E=KE+m_0{c}^2$

Understand the kinetic energy came into existence due to the electric field. Therefore, the kinetic energy will be equal to the energy associated with the charge particle due to the field.

Therefore,

$KE=qV$

Substitute $e$ for $q$

$KE=eV$

Combine the kinetic energy expression and the relativistic energy expression. Hence,

$E=eV+m_0{c}^2$

Also, the modified form of the relativistic energy is as follows:

$E^2=m_0{}^2{c}^4+p^2{c}^2$

Substitute $eV+m_0{c}^2$ for $E$ and solve

$\begin{array}{c}{\left(eV+m_0{c}^2\right)}^2=m_0{}^2{c}^4+p^2{c}^2\\ \left({\left(eV\right)}^2+{\left(m_0{c}^2\right)}^2+2m_0{c}^{2}eV\right)=m_0{}^2{c}^4+p^2{c}^2\\ {\left(eV\right)}^2+2m_0{c}^{2}eV=p^2{c}^2\\ {\left(\frac{eV}{c}\right)}^2+2m_{0}eV=p^2\end{array}$

Further solving, we get

$\begin{array}{c}{\left(\frac{eV}{c}\right)}^2+2m_{0}eV=p^2\\ p=\sqrt{{\left(\frac{eV}{c}\right)}^2+2m_{0}eV}\\ =\sqrt{2m_{0}eV\left(1+\frac{eV}{2m_0{c}^2}\right)}\end{array}$

Convert potential difference MV into V

$\begin{array}{c}1.0\text{ MV}=1.0\text{ MV}\left(\frac{{10}^6\text{ V}}{1\text{ MV}}\right)\\ =1\times {10}^6\text{ V}\end{array}$

Recall the relativistic effect, the expression for the De Broglie wavelength of the electron is

$\lambda =\frac{h}{\sqrt{2meV\left(1+\frac{eV}{2m{c}^2}\right)}}$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{ kg}$ for $m$, $1.6\times {10}^{-19}\text{ C}$ for $e$, $\left(2.998\times {10}^8\text{ m/s}\right)$ for $c$, and $1\times {10}^6\text{ V}$ for $V$

$\begin{array}{c}\lambda =\frac{6.63\times {10}^{-34}\text{ J}\cdot \text{s}}{\sqrt{2\left(9.11\times {10}^{-31}\text{ kg}\right)\left(1.6\times {10}^{-19}\text{ C}\right)\left(1\times {10}^6\text{ V}\right)\left(1+\frac{\left(1.6\times {10}^{-19}\text{ C}\right)\left(1\times {10}^6\text{ V}\right)}{2\left(9.11\times {10}^{-31}\text{ kg}\right){\left(2.998\times {10}^8\text{ m/s}\right)}^2}\right)}}\\ =8.7\times {10}^{-13}\text{ m}\end{array}$

Conclusion:

The De Broglie wavelength of the electron is $8.7\times {10}^{-13}\text{ m}$.