# If the proton is fired at a speed of 3.0x10^7m/s, what is its closest approach to the surface of...

## Question:

If the proton is fired at a speed of 3.0 {eq}\times {/eq} 10{eq}^7 {/eq}m/s, what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest.

## Conservation of Mechanical Energy

If only the conservative forces are acting on a system, the total mechanical energy of the system remains conserved, i.e. the sum of kinetic and potential energy remains constant, they may get interconverted into each other, where the conservative forces are those forces for which work done does not depends upon the path.

## Answer and Explanation:

**Data Given**

- Initial speed of the proton {eq}v = 3.0 \times 10^7 \ \rm m/s {/eq}

- Mass of the proton {eq}m = 1.67 \times 10^{-27} \ \rm kg {/eq}

Now let the proton is moving toward a nucleus of charge number **Z** as the proton approaches the nucleus its kinetic energy gets converted into electrostatic potential energy so using the conservation of mechanical energy

{eq}\begin{align} \frac{1}{2}mv^2 = \frac{k.e Ze}{r} \\ r = \frac{2Z ke^2}{mv^2} \end{align} {/eq}

This is the distance from the center of the nucleus to get the distance of the closest approach from the surface of the nucleus we must subtract the radius of the nucleus from the above-calculated distance i.e

{eq}\begin{align} r_{ca} = r-r_n \\ r_{ca} = \frac{2Z ke^2}{mv^2} - (A^{1/3}) r_0 \end{align} {/eq}

Where {eq}r_0 = 1.3 \ \rm fm {/eq}

Now let us calculate the distance of closest approach in case of simplest nucleus i.e. Hydrogen nucleus for which A = 1, and Z = 1 so

{eq}\begin{align} r_{ca} = \frac{2Z ke^2}{mv^2} - (Z^{1/3}) r_0 \\ r_{ca} = \frac{ 2 \times 1 \times 9 \times 10^9 \ \rm N.m^2 C^{-2} \times (1.6 \times 10^{-19} \ \rm C)^2}{1.67 \times 10^{-27} \ \rm kg \times (3.0 \times 10^7)^2 } - (1^{1/3}) \times 1.3 \times 10^{-15} \ \rm m \\ \color{ blue}{\boxed{ \ r_{ca} = -0.99 \times 10^{-15} \ \rm m \ }} \end{align} {/eq}

Here the negative sign indicates that the proton will enter into the nucleus and fuse to form a helium nucleus.

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from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6