Question:

The Delta baryon, \(\Delta\), is known as the lowest nucleon resonance that has a mass of 1232 MeV/c\(^2\) and a width of 120 MeV/c\(^2\). Its spin and isospin are equally 3/2. Find the lifetime of this particle.

Answer:

According to the uncertainty principle between time and energy, we have

\[  \Delta E \cdot \Delta t \geq \frac{\hbar}{2}  \]

The multiplication of uncertainties is greater than or equal to a half of the reduced Planck constant (or Dirac’s constant). Solve for the time and plug in numbers. In this unit system, the constant \(\hbar\) should be \(\hbar \times c\) = 197.33 MeV fm, where fm is 10\(^{-15}\) m. Therefore,

\[  \Delta t \sim \frac{\hbar c}{2\Delta E c} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{MeV \ fm}{MeV / c}} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{fm}{c}}  \]

\[  = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{10^{-15}m}{3.00 \times 10^8 m/s}} \]

\[  = 2.74 \times 10^{-24} \ \mathrm{s}  \]