Question

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?

Short answer

(a) The momentum of electron is $2.19\times {10}^{-26}\text{kg}·\text{m}/\text{s}$.

(b) 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) and case (b) are $1.25\times {10}^{-4}\%$ low and $40\%$low.

Step-by-step solution

Identification of given data

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

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

Momentum of electron

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.

Determination of momentum of electron

(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 $2.163\times {10}^{-26}\text{kg}·\text{m}/\text{s}$.

Determination of momentum of electron

(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 $3.64\times {10}^{-22}\text{kg}·\text{m}/\text{s}$.

Determination of errors in classical formula

(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.