Question

The neutral pion$\left(\tau {\tau }^0\right)$ 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 lifetime$\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathbf{}\mathit{s}$ of before decaying into two gamma-ray photons. Using the relationship$\mathit{E}\mathbf{=}\mathit{m}{\mathit{c}}^{\mathbf{2}}$ 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

$∆m=6.97\times {10}^{-36}kg=\left(2.9\times {10}^{-8}\right)m$

Step-by-step solution

Use Heisenberg’s Uncertainty principle to find uncertainty in energy

If the pion has a half-life of $8.4\times {10}^{-17}s$, we can take this as the uncertainty in the time $∆t=8.4\times {10}^{-17}s$.

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

$$\begin{gathered} ∆E∆t\ge \frac{\sout{h}}{2} \\ ∆E\ge \frac{\sout{h}}{2∆t} \end{gathered}$$

Now, we can get the minimum uncertainty in energy.

$$\begin{gathered} ∆E=\frac{1.055\times {10}^{-34}}{2\times 8.4\times {10}^{-17}} \\ ∆E=6.28\times {10}^{-19}J \end{gathered}$$

Use Energy-Mass relation to find uncertainty in mass

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

$$\begin{gathered} ∆m=\frac{E}{c^2} \\ ∆m=\frac{6.28\times {10}^{-19}}{{\left(3\times {10}^8\right)}^2} \\ ∆m=6.97\times {10}^{-36}kg \end{gathered}$$

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

$$\begin{gathered} ∆m=\frac{∆m}{m}m \\ ∆m=\frac{∆m}{264m_e} \\ ∆m=\frac{6.28\times {10}^{-36}}{264\times 9.11\times {10}^{-31}}m \\ ∆m=\left(2.9\times {10}^{-8}\right)m \end{gathered}$$