Answered: Consider an object of mass 11.1 kg. Now…

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 of mass 11.1 kg. Now imagine taking all the protons out of the object and stuffing them into a box. Also take all the electrons out of the object and stuff them in a second box.

You initially put the boxes of protons and electrons 39 m apart. Now you want to take the two boxes and move them farther apart, to a new distance of 70 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?

		1.45E+25 J
		2.89E+25 J
		5.78E+25 J
		1.16E+26 J

Expert Solution

Step 1

Given that

mass of an object is $M = 11.1 (k g)$

The initial position between the boxes of protons and electrons is $r_{1} = 39 (m)$

The final position between the boxes of protons and electrons is $r_{2} = 70 (m)$

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

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

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

If the number of electrons, protons, and neutrons are the same in the object (say $N$ ), then

$N (m_{p} + m_{n} + m_{e}) = M$

So the number of electrons, protons, and neutrons will be

$\begin{array}{rcl} N & = & \frac{M}{m_{p} + m_{n} + m_{e}} \\ = & \frac{11.1 (k g)}{1.67 \times 10^{- 27} (k g) + 1.67 \times 10^{- 27} (k g) + 9.1 \times 10^{- 31} (k g)} \\ = & 3.32 \times 10^{27} \end{array}$

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