Introduction To Health Physics
Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Chapter 3, Problem 3.35P
To determine

To calculate: The mass energy equivalence of a positron.

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A positive pion (π+, mass: m = 2.50 × 10-28 kg) at rest decays into a positive muon (μ+, mass: m₁ = 1.88 × 10-28 kg) and a neutrino. mμ (a) How much energy is released in the decay? You may assume the rest mass of the neutrino is negligible. Give your answer in MeV. (b) Give one reason why the neutrino is required in this decay. Briefly explain your answer.
A positron and an electron can annihilate each other and form 3 gammas.  Two gammas are detected.  One has energy of 225 keV, the other has 357 keV.  What is the energy of the 3rd gamma?
Current Attempt in Progress Find the energy (in MeV) released when a decay converts 88X226 (atomic mass = 226.02540 u) into 86Y222 (atomic mass = 222.01757 u). The atomic mass of an a particle is 4.002603 u. Number i Save for Later Parent nucleus N Z+ Units Daughter nucleus N-20 Z-2+ a particle (helium nucleus) + 2 2+ Attempts: 0 of 1 used Submit Answer
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