# A proton has been accelerated from rest through a potential difference of -1350 V. a) What is...

## Question:

A proton has been accelerated from rest through a potential difference of -1350 V.

a) What is the proton's kinetic energy, in electron volts?

b) What is the proton's kinetic energy, in joules?

c) What is the proton's speed?

## The change in Kinetic Energy and Electric Potential Difference:

The change in kinetic energy of a charged particle when it passes through a potential difference {eq}\Delta V {/eq} is given by the equation;

{eq}\Delta (K.E.) = q \Delta V {/eq}

where, {eq}q {/eq} is the charge of the charge particle and {eq}\Delta (K.E.) {/eq} is the change in the kinetic energy.

## Answer and Explanation:

- The potential difference through which the proton is accelerated is {eq}\Delta V = -1350\ V {/eq}

- The proton's charge is {eq}q =e= 1.6\times 10^{-19}\ C {/eq}

- The mass of proton is {eq}m = 1.672\times 10^{-27}\ kg {/eq}

**Part (a)**

The kinetic energy of the proton is given by,

{eq}\begin{align} K.E. &=q|\Delta V|\\ &= (e)\times (1350\ V)\\ &= 1350\ eV\\ \end{align} {/eq}

**Part (b)**

The kinetic energy of the proton in joules is

{eq}\begin{align} K.E. &= 1350\ eV\\ &= 1350 \times (1.6\times 10^{-19})\ J \\ &= 2.160\times 10^{-16}\ J\\ \end{align} {/eq}

**Part (c)**

The kinetic energy of the proton can be written as

{eq}\begin{align} K.E. &= \dfrac{1}{2}mv^2\\ \implies v&= \sqrt{\dfrac{2K.E.}{m}}\\ &= \sqrt{\dfrac{2\times 2.160\times 10^{-16}\ J }{ 1.672\times 10^{-27}\ kg}}\\ &=5.083\times 10^{5}\ m/s\\ \end{align} {/eq}

**Thus, the speed of the proton is {eq}v = 5.083\times 10^{5}\ m/s {/eq}**

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Chapter 4 / Lesson 14