# Two protons are released from rest when they are 0.750 nm apart. What is the maximum speed they...

## Question:

Two protons are released from rest when they are 0.750 nm apart. What is the maximum speed they will reach?

## Conservation of Energy

In mechanics, when a process has occurred, then it initial to final energy is conserved. Energy only can be transfer from one state to another state, and it can't be made by anyone or destroy by any method. According to the law of conservation of energy, the total summation of the initial energy is equal to the summation of the final energy.

## Answer and Explanation:

Based on the problem, we know that the distance between two protons is {eq}x = 0.750\;{\rm{nm}}. {/eq}

Recall that the expression for the conservation of mass can be calculated as follow:

$$\begin{align*} \dfrac{1}{2}M{V^2} + \dfrac{1}{2}M{V^2} &= k\dfrac{{{Q_1}{Q_2}}}{x}\\ M{V^2}& = k\dfrac{{{Q_1}{Q_2}}}{x}\\ V &= \sqrt {k\dfrac{{{Q_1}{Q_2}}}{{Mx}}} \end{align*} $$

where the mass of proton is {eq}M = 1.67 \times {10^{ - 27}}\;{\rm{kg}} {/eq} , {eq}V {/eq} is velocity and the charges of protons are {eq}{Q_1},{Q_2} = 1.6 \times {10^{ - 9}}\;{\rm{C}} {/eq} and {eq}k = 9 \times {10^9}\;{{{\rm{N}}{{\rm{m}}^2}} {\left/ {\vphantom {{{\rm{N}}{{\rm{m}}^2}} {{{\rm{C}}^2}}}} \right. } {{{\rm{C}}^2}}} {/eq} is the Coulomb's constant.

Substitute the known values in above equation to get maximum velocity.

$$\begin{align*} V &= \sqrt {\left( {9 \times {{10}^9}} \right)\dfrac{{\left( {1.6 \times {{10}^{ - 9}}} \right)\left( {1.6 \times {{10}^{ - 9}}} \right)}}{{\left( {1.67 \times {{10}^{ - 27}}} \right)\left( {0.750\times {{10}^{ - 9}}} \right)}}} \\ &= 1.356 \times {10^{14}}\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} $$

**Thus, the maximum speed is {eq}1.356 \times {10^{14}}\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}. {/eq} **

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from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6