A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly off in oppos

Question

A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly off in opposite directions, each with speed 0.827c relative to an inertial reference frame S. What is the total energy of particle A?

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Calantha 4 years 2021-09-01T21:33:36+00:00 1 Answers 31 views 0

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  1. thachthao
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    2021-09-01T21:35:05+00:00
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    Answer:

    E = 389 MeV

    Explanation:

    The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

    Total Energy of A = E = Rest Mass Energy + Relative Energy

    Using Einstein’s Equation: E = mc²

    E = m₀c² + mc²

    From Einstein’s Special Theory of Relativity, we know that:

    m = m₀/[√(1-v²/c²)]

    Therefore,

    E = m₀c² + m₀c²/[√(1-v²/c²)]

    E = m₀c²[1 + 1/√(1-v²/c²)]

    where,

    m₀c² = rest mass energy = 140 MeV

    v = relative speed = 0.827 c

    Therefore,

    E = (140 MeV)[1 + 1/√(1 – (0.827c)²/c²)]

    E = (140 MeV)(2.78)

    E = 389 MeV

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