Question

An electron is projected with an initial speed vi = 4.60 × 105 m/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 3vi.

Answer 1

Answer:

$2.99\times 10^{-19}\ m$

Explanation:

Given:

• $u$ = initial velocity of the electron = $4.60\times 10^5\ m/s$

• $v$ = final velocity of the electron = $3u$

• $x$ = initial position of the electron from the proton = very distant =  $\infty$

Assume:

• $m$ = mass of an electron = $9.1\times10^{-31}\ kg$

• $e$ = magnitude of charge on an electron = $1.6\times10^{-19}\ C$

• $p$ = magnitude of charge on an proton = $1.6\times10^{-19}\ C$

• $k$ = Boltzmann constant = $9\times 10^9\ Nm^2/C^2$

• $y$ = final position of the electron from the proton

• $\Delta K$ = change in kinetic energy of the electron

• $W$ = work done by the electrostatic force

• $F$ = electrostatic force

• $r$ = instantaneous distance of the electron from the proton

Let us first calculate the work done by the electrostatic force.

$W=\int Fdr\\\Rightarrow W = \int \dfrac{kep}{r^2}dr\\\Rightarrow W = kep\int \dfrac{1}{r^2}dr\\\Rightarrow W = kep\left | \dfrac{1}{r} \right |_{y}^{x}\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{\infty} \right )\\\Rightarrow W =\dfrac{kep}{x}$

Using the principle of the work-energy theorem,

As only the electrostatic force is assumed to act between the two charges, the kinetic energy change of the electron will be equal to the work done by the electrostatic force on the electron due to proton.

$\therefore \Delta K = W\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m((3u)^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m(8u^2)= \dfrac{kep}{x}\\\Rightarrow x= \dfrac{2kep}{8mu^2}\\\Rightarrow x= \dfrac{2\times 9\times 10^9\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{8\times 9.1\times10^{-31}\times (4.60\times 10^5)^2}\\\Rightarrow x=2.99\times 10^{-10}\ m\leq$

Hence, the electron is at a distance of $2.99\times 10^{-10}\ m$ when the electron instantaneously has speed of three times the initial speed.