Question

Express in $eV$(or $keV$or $MeV$if more appropriate):

a. The kinetic energy of an electron moving with a speed of $5.0\times {10}^{6}m/s$.

b. The potential energy of an electron and a proton $0.10nm$apart.

c. The kinetic energy of a proton that has accelerated from rest through a potential difference of $5000V$.

Short answer

a. The kinetic energy of an electron is$E_k=71.17eV$

b. The potential energy of an electron and a proton $0.10nm$apart is$U=-14.42eV$.

c. The kinetic energy of a proton is$E_k=5keV$.

Step-by-step solution

Part (a) step 1: Given information

We have given that the express in $eV$.

We need to find kinetic energy of an electron moving with a speed of$5.0\times {10}^{6}m/s$.

Part (a) step 2: Simplify

The expression for kinetic energy

$$\begin{gathered} E_k=\frac{1}{2}m{v}^2 \\ {E}_k=\frac{1}{2}(9.11)\left({10}^{-31}\right)(5{\left({10}^6\right))}^2{E}_k=1.13875({10}^{-17})J \end{gathered}$$

Part (b) step 1: Given information 

We have given that the express in$eV$

We need to find potential energy of an electron and a proton $0.10nm$apart.

Part (b) step 2: Simplify

Using formula for potential energy :

$$\begin{gathered} U=\frac{q_1{q}_2}{4\mathrm{\pi }{\in }_0\mathrm{r}} \\ U=\frac{e(-e)}{4\mathrm{\pi }{\in }_0(0.1)\left({10}^{-9}\right)} \\ U=-2.307\left({10}^{-18}\right)J=-14.42eV \end{gathered}$$

Part (c) step 1: Given information

We have given that the express in $eV$

We need to find the kinetic energy of a proton that has accelerated from rest through a potential difference of$5000V$.

Part (c) step 2: Simplify

The expression for kinetic energy when charge accelerating with a potential difference $∆V$:

$$\begin{gathered} E_k=q∆V \\ {E}_k=e∆V \\ {E}_k=e\left(5000\right) \\ {E}_k=5keV \end{gathered}$$