Determine the momentum of an electron moving (a) at speed $\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$m/s (about three times escape velocity) and (b) at speed $\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$m/s. (c) In each case by how much is the classical formula in error?

The speed of electron is $v_1=2.4\times {10}^4\text{m}/\text{s}$

The speed of electron is $v_2=2.4\times {10}^8\text{m}/\text{s}$

The momentum of a particle is the effect due to the mass and speed of the particle. The variation in this effect with time is equal to the force of the particle.

(a)

The momentum of electron is given as:

$P_1=\frac{m{v}_1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}$

Here is the mass of electron and its value is $9.01\times {10}^{-31}\text{kg}$, c is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$.

Substitute all the values in the equation.

$$\begin{gathered} P_1=\frac{\left(9.01\times {10}^{-31}\text{kg}\right)\left(2.4\times {10}^4\text{m}/\text{s}\right)}{\sqrt{1-{\left(\frac{2.4\times {10}^4\text{m}/\text{s}}{3\times {10}^8\text{m}/\text{s}}\right)}^2}} \\ {P}_1=2.163\times {10}^{-26}\text{kg}·\text{m}/\text{s} \end{gathered}$$

Therefore, the momentum of electron is $2.163\times {10}^{-26}\text{kg}·\text{m}/\text{s}$.

(b)

The momentum of electron is given as:

$P_2=\frac{m{v}_2}{\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}}$

Substitute all the values in the equation.

$$\begin{gathered} P_2=\frac{\left(9.01\times {10}^{-31}\text{kg}\right)\left(2.4\times {10}^8\text{m}/\text{s}\right)}{\sqrt{1-{\left(\frac{2.4\times {10}^8\text{m}/\text{s}}{3\times {10}^8\text{m}/\text{s}}\right)}^2}} \\ {P}_2=3.64\times {10}^{-22}\text{kg}·\text{m}/\text{s} \end{gathered}$$

Therefore, The momentum of electron is $3.64\times {10}^{-22}\text{kg}·\text{m}/\text{s}$.

(c)

The error in classical formula for case (a) is given as:

$x_1=\frac{P_1-m{v}_1}{P_1}$

Substitute all the values in the equation.

$$\begin{gathered} x_1=\frac{2.163\times {10}^{-26}\text{kg}·\text{m}/\text{s}-\left(9.01\times {10}^{-31}\text{kg}\right)\left(2.4\times {10}^4\text{m}/\text{s}\right)}{2.163\times {10}^{-26}\text{kg}·\text{m}/\text{s}} \\ {x}_1=1.25\times {10}^{-2} \\ {x}_1=1.25\times {10}^{-4}\% \end{gathered}$$

The error in classical formula for case (b) is given as:

$x_2=\frac{P_2-m{v}_2}{P_2}$

Substitute all the values in the equation.

$$\begin{gathered} x_2=\frac{3.60\times {10}^{-22}\text{kg}·\text{m}/\text{s}-\left(9.01\times {10}^{-31}\text{kg}\right)\left(2.4\times {10}^8\text{m}/\text{s}\right)}{3.60\times {10}^{-22}\text{kg}·\text{m}/\text{s}} \\ {x}_2=0.4 \\ {x}_2=40\% \end{gathered}$$

Therefore, the error in classical formula for case (a) and case (b) are $1.25\times {10}^{-4}\%$ low and $40\%$low.