Question

An electron is projected with an initial speed of $\mathbf{3}\mathbf{.2}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{\text{m/s}}$directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

Short answer

The distance from the proton is$1.645\times {10}^{-9}\text{m}$ where the speed of the electron instantaneously equal to twice the initial value.

Step-by-step solution

 Step 1: The given data 

1. Mass of the electron, $\mathrm{m}=9.11\times {10}^{-31}\text{kg}$
2. Initial speed of the electron, ${\mathrm{v}}_{\mathrm{i}}=3.2\times {10}^5\text{m/s}$

Understanding the concept of energy

Using the concept of energy, we calculate both the initial and the final kinetic energies of the system by using the given data in the formula. Now, the difference between these energies is compared to the work done by the system, this determines the final velocity of the particle.

Formulae:

The kinetic energy of the electron before colliding with proton is given by,

$\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$ (i)

Here, m is the mass of the electron and $\mathrm{v}$is the velocity.

The work done by the proton on the electron, $\mathrm{W}=\mathrm{qV}$ (ii)

The potential energy of the system, $\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\mathrm{r}}$ (iii)

Calculation of the distance from the proton 

The value of the kinetic energy is given using the given data in equation (i) as follows:

$\begin{array}{c}\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}=\frac{1}{2}(9.11\times {10}^{-31}\text{kg}){(3.2\times {10}^5\text{m/s})}^2\\ =46.64\times {10}^{-21}\text{J}\end{array}$

When electron acquires a velocity double the value of initial velocity. So, the final kinetic energy is given using equation (i) as follows:

$\begin{array}{c}\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}=\frac{2}{2}(9.11\times {10}^{-31}\text{kg}){(3.2\times {10}^5\text{m/s})}^2\\ =186.57\times {10}^{-21}\text{J}\end{array}$

Substituting the values $1.6\times {10}^{-19}\mathrm{C}$for $\mathrm{q}$and $8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2$for ${\mathrm{\epsilon }}_0$and the expression for work done to the difference in initial and final kinetic energies is given using the given data and equations (iii) in (ii) as follows:

$\begin{array}{c}\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}-\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}=\mathrm{W}\\ (186.57\times {10}^{-21}\text{J})-(46.64\times {10}^{-21}\text{J})=\frac{1}{4\mathrm{\pi }(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/Nm}}^{\text{2}})}\frac{{(1.6\times {10}^{-19}\text{C})}^2}{\mathrm{r}}\\ 139.93\times {10}^{-21}\text{J}=\frac{1}{111.156\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/Nm}}^{\text{2}}}\frac{2.56\times {10}^{-38}{\text{C}}^{\text{2}}}{\mathrm{r}}\\ \mathrm{r}=1.645\times {10}^{-9}\text{m}\end{array}$

Hence, the distance of electron from proton is $1.645\times {10}^{-9}\text{m}$.