The explosive yield of the atomic bomb dropped on Hiroshima near the end of World War II was approximately 15.0 kilotons of TNT. One kiloton is about \(4.18 \cdot 10^{12} \mathrm{~J}\) of energy. Find the amount of mass that was converted into energy in this bomb.

Short Answer

Expert verified
Answer: Approximately \(6.97 \cdot 10^{-4}\) kg of mass was converted into energy in the atomic bomb dropped on Hiroshima.

Step by step solution

01

Understand the given information

We are given the explosive yield of the atomic bomb dropped on Hiroshima as 15.0 kilotons of TNT, which is equivalent to 15.0 \(\times 4.18 \cdot 10^{12} \mathrm{~J}\) of energy. We will use this information to find the mass converted into energy.
02

Convert the explosive yield in kilotons to Joules

To find the energy in Joules, we need to multiply the explosive yield in kilotons by the conversion rate provided. That is: Energy (J) = 15.0 kilotons × (\(4.18 \cdot 10^{12}\) J/kiloton) thus, Energy (J) = \(6.27 \cdot 10^{13}\) J
03

Use Einstein's equation to find the mass

Einstein's equation states that the energy of an object is equal to its mass multiplied by the speed of light squared (\(E=mc^2\)). We have the energy (E) and need to find the mass (m). The speed of light (c) is approximately \(3.0 \cdot 10^8\ \mathrm{m/s}\) Rearranging the equation to solve for mass, we get: m = \(\frac{E}{c^2}\) Substituting the values we have m = \(\frac{6.27 \cdot 10^{13} \mathrm{~J}}{(3.0 \cdot 10^8\ \mathrm{m/s})^2}\)
04

Calculate the mass

Now calculate the mass by dividing the energy by the speed of light squared: m = \(\frac{6.27 \cdot 10^{13} \mathrm{~J}}{(3.0 \cdot 10^8\ \mathrm{m/s})^2}\) m = \(\frac{6.27 \cdot 10^{13} \mathrm{~J}}{9.0 \cdot 10^{16}\ \mathrm{m^2/s^2}}\) m ≈ \(6.97 \cdot 10^{-4}\) kg So, about \(6.97 \cdot 10^{-4}\) kg of mass was converted into energy in the atomic bomb dropped on Hiroshima.

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.

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

Consider a one-dimensional collision at relativistic speeds between two particles with masses \(m_{1}\) and \(m_{2}\). Particle 1 is initially moving with a speed of \(0.700 c\) and collides with particle \(2,\) which is initially at rest. After the collision, particle 1 recoils with speed \(0.500 c\), while particle 2 starts moving with a speed of \(0.200 c\). What is the ratio \(m_{2} / m_{1} ?\)

A proton with a momentum of \(3.0 \mathrm{GeV} / \mathrm{c}\) is moving with what velocity relative to the observer? a) \(0.31 c\) c) \(0.91 c\) e) \(3.2 c\) b) \(0.33 c\) d) \(0.95 c\)

In an elementary-particle experiment, a particle of mass \(m\) is fired, with momentum \(m c\), at a target particle of mass \(2 \sqrt{2} m\). The two particles form a single new particle (completely inelastic collision). Find: a) the speed of the projectile before the collision b) the mass of the new particle c) the speed of the new particle after the collision

Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, you see that the particle is, in fact, moving toward the wire.)

The hot filament of the electron gun in a cathode ray tube releases electrons with nearly zero kinetic energy. The electrons are next accelerated under a potential difference of \(5.00 \mathrm{kV}\), before being steered toward the phosphor on the screen of the tube. a) Calculate the kinetic energy acquired by the electron under this accelerating potential difference. b) Is the electron moving at relativistic speed? c) What is the electron's total energy and momentum? (Give both values, relativistic and nonrelativistic, for both quantities.)

See all solutions

Recommended explanations on Physics Textbooks

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