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$m_{∧}c_{2}=1115.6MeV$ $m_{p}c_{2}=938.3MeV$

$mπc_{2}=139.6MeV$

The Q value of the reaction, representing the energy output, is the difference between starting rest energy and final rest energy, and is the kinetic energy of the products:

$Q=1115.6MeV−938.3MeV−139.6MeV=37.7MeV$

(b) The original kinetic energy is zero in the process considered here, so the whole Q becomes the kinetic energy of the products

$K_{p}+K_{π}=37.7MeV$

(c) The lambda particle is at rest. Its momentum is zero. System momentum is conserved in the decay, so the total vector the momentum of the proton and the pion must be zero.

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