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Q23P

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Fundamentals Of Physics
Found in: Page 1364

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Short Answer

Consider the decay Λ0p+πwith theΛ0 at rest. (a) Calculate the disintegration energy. What is the kinetic energy of (b) the proton and (c) the pion?

(a) The disintegration energy is.37.7 Mev

(b) The kinetic energy of proton is.5.35 Mev

(c) The kinetic energy of pion is.32.4 Mev

See the step by step solution

Step by Step Solution

Step 1: (a) Evaluate disintegration energy.

The given reaction is written as:

Λ0p+π

Solve the disintegration energy as follows:

Q=Δmc2=(mΛ0-mp-mπ-)c2

Substitute the values from the standard data.

Q=(1115.6-938.3-139.6)Mev=37.7Mev

Thus, the disintegration energy is.37.7Mev

Step 2: (b) Evaluate the kinetic energy of proton. 

Solve the kinetic energy of proton in the given reaction as follows;

Kp=(EΛ-Ep)2-Eπ22EΛ

Substitute the values from the standard data

Kp=(1115.6-938.3)2-(139.6)22(1115.6) Mev=5.35 Mev

Thus, the kinetic energy of proton is .5.35Mev

Step 3: (c) Evaluate the kinetic energy of pion.

Solve the kinetic energy of proton in the given reaction as follows;

Kπ=Q-Kp=(37.7-5.35) Mev=32.4​ Mev

Thus, the kinetic energy of pion is.32.4Mev

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