Carbon-14 will emit a $\beta$ particle with an energy of 0.1565 MeV. What is this energy in joules? A) 1.0 × 10^-24 J B) 2.5 × 10^-20 J C) 1.0 × 10^-18 J D) 2.5 × 10^-14 J E) none of the above

Question

Quizplus · Accepted Answer

The energy of a particle can be calculated using the equation E = mc^2, where E is the energy in joules, m is the mass of the particle in kilograms, and c is the speed of light in meters per second. In this case, the mass of the particle is 1.6605 × 10^-27 kg and the speed of light is 2.998 × 10^8 m/s. Plugging these values into the equation, we get E = (1.6605 × 10^-27 kg) * (2.998 × 10^8 m/s)^2 = 2.5 × 10^-14 J.

Carbon-14 Will Emit A β\betaβ Particle with an Energy of 0

Carbon-14 Will Emit A $\beta$ Particle with an Energy of 0