Question

An electron is projected with an initial speed${\mathit{v}}_{\mathbf{i}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.2}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\frac{\mathbf{\text{m}}}{\mathbf{\text{s}}}$directly toward a very distant proton that is at rest. Because the proton mass is large relative to the electron mass, assume that the proton remains at rest. By calculating the work done on the electron by the electrostatic force, determine the distance between the two particles when the electron instantaneously has speed$\mathbf{2}{\mathit{v}}_{\mathbf{i}}$.

Short answer

The distance between the particles is $r=1.64\times {10}^{-9}\text{\hspace{0.17em}m}$.

Step-by-step solution

 Step 1: Given

The initial speed of the particle is$v_i=3.2\times {10}^5\frac{\text{m}}{\text{s}}$.

Determine the concept

According to the principle of conservation of energy, energy is neither creatednor destroyed.It may only transform from one type to another

Calculate the distance between the two particles when the electron instantaneously has speed2vi

Since the total energy is conserved the required relation is:

$\frac{1}{2}{m}_e{v}_i^2=\frac{1}{2}{m}_e{v}_f^2-\frac{k{e}^2}{r_f}$

Here,$r_f$is the distance between the electron and the proton.

For$v_f=2v_i$, and for$r_f$solve as:

$\begin{array}{c}{r}_f=\frac{2k{e}^2}{m_e(v_f^2-v_i^2)}\\ r_f=\frac{2k{e}^2}{3m_e{v}_i^2}\end{array}$

$\begin{array}{c}{r}_f=\frac{2(9\times {10}^9\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}}){(1.60\times {10}^{-19}\text{\hspace{0.17em}C})}^2}{3(9.11\times {10}^{-31}\text{\hspace{0.17em}kg}){(3.20\times {10}^5\frac{\text{m}}{\text{s}})}^2}\\ =1.64\times {10}^{-9}\text{m}\end{array}$

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