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Chapter 6: Q74P (page 1318)

The neutral pion(蟿蟿0) is an unstable particle produced in high-energy particle collisions. Its mass is about 264 times that of the electron, and it exists for an average lifetime8.4脳10-17s of before decaying into two gamma-ray photons. Using the relationshipE=mc2 between rest mass and energy, find the uncertainty in the mass of the particle and express it as a fraction of the mass.

Short Answer

Expert verified

鈭�m=6.97脳10-36kg=2.9脳10-8m

Step by step solution

01

Use Heisenberg’s Uncertainty principle to find uncertainty in energy

If the pion has a half-life of 8.4脳10-17s, we can take this as the uncertainty in the time 鈭�t=8.4脳10-17s.

From the Heisenberg鈥檚 uncertainty principle, the uncertainty in the energy of a state that is occupied is given by:

鈭�E鈭�t鈮�h2鈭�E鈮�h2鈭�t

Now, we can get the minimum uncertainty in energy.

鈭�E=1.055脳10-342脳8.4脳10-17鈭�E=6.28脳10-19J

02

Use Energy-Mass relation to find uncertainty in mass

The uncertainty in energy is due to uncertainty in mass which can be expressed as .

鈭�m=Ec2鈭�m=6.28脳10-193脳1082鈭�m=6.97脳10-36kg

The uncertainty in the mass of the pion expressed as a fraction of its mass is

鈭�m=鈭�mmm鈭�m=鈭�m264me鈭�m=6.28脳10-36264脳9.11脳10-31m鈭�m=2.9脳10-8m

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