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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}


Learn more about this topic:

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Kinetic Energy: Examples & Definition

from General Studies Science: Help & Review

Chapter 4 / Lesson 14
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