Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 70

Alpha decay is nuclear decay in which a helium nucleus is emitted. If the helium nucleus has a mass of \(6.80 \times 10^{-27} \mathrm{kg}\) and is given \(5.00 \mathrm{MeV}\) of kinetic energy, what is its velocity?

Short Answer

Expert verified
The velocity of the emitted helium nucleus during alpha decay is approximately \(4.85 \times 10^7\, \mathrm{m/s}\).

Step by step solution

01

Write down the formula for kinetic energy

The formula for kinetic energy is given by: \[KE = \frac{1}{2}mv^2\], where KE is the kinetic energy, m is the mass, and v is the velocity.
02

Convert the given kinetic energy to Joules

We are given the kinetic energy in MeV (Mega-electron volts), and we need to convert it to Joules (SI unit). We can use the following conversion factor: \[1\,\mathrm{eV} = 1.602 \times 10^{-19}\, \mathrm{J}\] So, for 5.00 MeV, we have: \[5.00\,\mathrm{MeV} \times 10^6\, \mathrm{eV/MeV} \times 1.602 \times 10^{-19}\, \mathrm{J/eV} = 8.01 \times 10^{-13}\, \mathrm{J}\]
03

Plug in the values into the kinetic energy formula and solve for the velocity

Now we have the mass, m = \(6.80 \times 10^{-27}\, \mathrm{kg}\), and the kinetic energy, KE = \(8.01 \times 10^{-13}\, \mathrm{J}\). We can plug these values into the kinetic energy formula and solve for the velocity, v: \[8.01 \times 10^{-13}\, \mathrm{J} = \frac{1}{2}(6.80 \times 10^{-27}\, \mathrm{kg})v^2\] Now, we solve for v: \[v^2= \frac{2 \times (8.01 \times 10^{-13}\, \mathrm{J})}{6.80 \times 10^{-27}\, \mathrm{kg}}\] \[v^2 = 2.35 \times 10^{14}\, \mathrm{m^2/s^2}\] Taking the square root of both sides, we get the velocity: \[v = \sqrt{2.35 \times 10^{14}\, \mathrm{m^2/s^2}}\]
04

Calculate the velocity

Now we can calculate the velocity: \[v = \sqrt{2.35 \times 10^{14}\, \mathrm{m^2/s^2}} = 4.85 \times 10^7\, \mathrm{m/s}\] Therefore, the velocity of the emitted helium nucleus is approximately \(4.85 \times 10^7\, \mathrm{m/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the rest energy of an electron, given its mass is \(9.11 \times 10^{-31} \mathrm{kg} ?\) Give your answer in joules and \(\mathrm{MeV}\)

A highway patrol officer uses a device that measures the speed of vehicles by bouncing radar off them and measuring the Doppler shift. The outgoing radar has a frequency of \(100 \mathrm{GHz}\) and the returning echo has a frequency \(15.0 \mathrm{kHz}\) higher. What is the velocity of the vehicle? Note that there are two Doppler shifts in echoes. Be certain not to round off until the end of the problem, because the effect is small.

To whom does the elapsed time for a process seem to be longer, an observer moving relative to the process or an observer moving with the process? Which observer measures the interval of proper time?

If a spaceship is approaching the Earth at \(0.100 c\) and a message capsule is sent toward it at \(0.100 c\) relative to Earth, what is the speed of the capsule relative to the ship?

Explain the meaning of the terms "red shift" and "blue shift" as they relate to the relativistic Doppler effect.
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Measurements

Read Explanation

Fluids

Read Explanation

Torque and Rotational Motion

Read Explanation

Classical Mechanics

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free