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Fundamentals Of Physics
Found in: Page 1364

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Short Answer

The spin- 32Σ*0baryon (see table in Problem 24) has a rest energy of (1385MeVwith an intrinsic uncertainty ignored here); the spin12Σ0- baryon has a rest energy of.1192.5MeVIf each of these particles has a kinetic energy of 1000MeV,

(a) which is moving faster and

(b) by how much?

(a) The spin- 12Σ0baryon is moving faster.

(b) It is moving faster by.7.5×106ms

See the step by step solution

Step by Step Solution

Step 1: Given data

Rest energy of the spin- 32Σ0 baryon

E0=1385 MeV

Rest energy of the spin- 12Σ0baryon

E0=1192.5 MeV

Kinetic energy of the two particles

K=1000 MeV

Step 2:Determine the formula for the relativistic energy.

The relativistic kinetic energy of a particle moving with velocity v and having rest energy E0 is:

K=(11-v2c2-1)E0 ..... (I)

Here,c is the speed of light in vacuum with value


Step 3: Determinethe particle with greater velocity


Let the velocity of the spin-32Σ0 baryonbe v . Then from equation (I)


Substitute the values and solve as:

1000 MeV=(11v2c21)×1385 MeV11v2c2=1.722v2c2=0.663v=0.814c

Let the velocity of the spin-12Σ0 baryon be .v Then from equation (I)


Substitute the values and solve as:

1000 MeV=(11v2c21)×1192.5 MeV11v2c2=1.839v2c2=0.704v=0.839c

Compare the two velocities:


Thus, the speed of the spin- 12Σ0is greater.

Step 4: Determine the difference in velocities of the two particles


The speed of the spin32Σ*0- baryon is


The speed of the spin-12Σ0 baryon is


The difference in their speeds is calculated as:


Thus, the difference is.7.5×106ms


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