Answered: Your object has mass 39.3 kg, and youve…

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

Your object has mass 39.3 kg, and youve separated the boxes of protons and electrons by a distance of 23 m. If you were to release the clump of protons from their box, how much acceleration would the clump of protons undergo due to its attraction to the clump of electrons in the other box?

	A.	1.54E+24 m/s^2
	B.	3.07E+24 m/s^2
	C.	6.14E+24 m/s^2
	D.	1.23E+25 m/s^2

Expert Solution

Step 1

Given that,

The mass of an object is $M = 39.3 (k g)$

The separation between the boxes of proton and electron is $r = 23 (m)$

Now since 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 $m_{n} = 1.67 \times 10^{- 27} (k g)$

Let the object have an equal number of protons, electrons, and neutrons (say $n$ )

Then

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

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

$\begin{array}{rcl} n & = & \frac{M}{m_{p} + m_{n} + m_{e}} \\ = & \frac{39.3 (k g)}{1.67 \times 10^{- 27} (k g) + 1.67 \times 10^{- 27} (k g) + 9.1 \times 10^{- 31} (k g)} \\ = & 11.763 \times 10^{27} \end{array}$

So the number of protons and electrons, in the box is $\begin{array}{rcl} n & = & 11.763 \times 10^{27} \end{array}$

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