Answered: Consider an object has a mass of 43.4…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

Consider an object has a mass of 43.4 kg. You separate the protons and electrons of the object into two boxes. You take all the protons out of the object and stuff them into a box and also take all the electrons out of the object and stuff them into a second box.

You initially put the boxes of protons and electrons 45 m apart. Now you want to take the two boxes and move them farther apart, to a new distance of 81 m apart. How much work (that is, energy) would you have to expend to pull the boxes that much farther apart, fighting against the electrostatic attraction between them?

		1.92E+26 J
		3.84E+26 J
		7.68E+26 J
		1.54E+27 J

Expert Solution

Step 1

Given that,

the mass of an object is $M = 43.4 (k g)$

The initial separation between the proton and electron box is $r_{o} = 45 (m)$

The final separation between the proton and electron box is $r = 81 (m)$

Since the mass of the electron is $m_{e} = 9.1 \times 10^{- 31} (k g)$

the mass of the proton is $m_{p} = 1.67 \times 10^{- 27} (k g)$

the mass of the neutron is $m_{n} = 1.67 \times 10^{- 27} (k g)$

Considering the object has an equal number of electrons, protons, and neutrons, then

$\begin{array}{rcl} (m_{e} + m_{p} + m_{n}) N & = & M \\ N & = & \frac{M}{(m_{e} + m_{p} + m_{n})} \end{array}$

Putting the values

$N = \frac{43.4 (k g)}{(9.1 \times 10^{- 31} (k g) + 1.67 \times 10^{- 27} (k g) + 1.67 \times 10^{- 27} (k g))} N = 12.99 \times 10^{27}$

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