91Ó°ÊÓ

Total energy content of a particle is composed of two types of energies, one is the rest mass energy of the particle and the other is the kinetic energy of the particle.

a)

The quark composition for\({{\rm{\Delta }}^{\rm{ + }}}\)is given in table\({\rm{33}}{\rm{.4}}\)as:

\({{\rm{\Delta }}^{\rm{ + }}}{\rm{ = [uud]}}\)

From the same table, we can observe that the proton's quark makeup is: i

\({\rm{p = [u u d] }}\)

As they have the same quark makeup, so one of them must be the excited state of the other. But, which of these represents the excited state? Because the proton is a stable particle, the particle\({{\rm{\Delta }}^{\rm{ + }}}\)is the proton's excited state, as seen in the following decay:

\({\Delta ^ + } \to p + \gamma \)

b)

According to the Heisenberg uncertainty principle, energy uncertainty is connected to temporal uncertainty in the following way:

\(\Delta E\Delta t \le \frac{h}{{4\pi }}\)

Where,\({\rm{h}}\)is Planck's constant and it is equal to\(h = 6.626 \times {10^{ - 34}}\;{\rm{J}}{\rm{.s}}\)Thus, the approximate lifetime for the lifetime is given by:

\(\Delta t \simeq \frac{h}{{4\pi \Delta E}}\)

The uncertainty in energy of the\({{\rm{\Delta }}^{\rm{ + }}}\)decay is:

\(\begin{array}{c}\Delta E = 100\;{\rm{Mev}}\\ = \left( {100 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}\;{\rm{J}}} \right)\end{array}\)

Therefore, the solution is

\(\begin{array}{c}\Delta t \simeq \frac{h}{{4\pi \Delta E}}\\ \simeq \frac{{6.626 \times {{10}^{ - 34}}\;{\rm{J}}{\rm{.s}}}}{{4\pi \times \left( {100 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}} \right)\;{\rm{J}}}}\\ \simeq 3.296 \times {10^{ - 24}}\;{\rm{s}}\end{array}\)

The lifetime of \({{\rm{\Delta }}^{\rm{ + }}}\) is \(3.296 \times {10^{ - 24}}\;{\rm{s}}\).

In this section, we need to figure out what forces cause the particle \({{\rm{\Delta }}^{\rm{ + }}}\)to decay. A number of criteria may be used to answer this question. For starters, as a baryon, \({{\rm{\Delta }}^{\rm{ + }}}\)is susceptible to the strong nuclear force, which causes decay. The weak force does not play a role in this decay since the quark flavor does not change. As a result, the powerful force bears responsibility for the decay.