Uncertainty principle: Time and energy with Delta baryon

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
\[
\Delta E \cdot \Delta t \geqq \frac{\hbar}{2}
\]
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,
\[
\Delta t \sim \frac{\hbar c}{2\Delta E c} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{MeV \ fm}{MeV / c}} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{fm}{c}}  \\
= \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{10^{-15}m}{3.00 \times 10^8 m/s}} \\
= 2.74 \times 10^{-24} \ \mathrm{s}
\]

Powered by Hirophysics.com