Question

Estimate the range of the force mediated by an \({\omega ^0}\)g meson that has mass \(783\;MeV/{c^2}\).

Short answer

The range of the force is \(1.26 \times 1{0^{ - 16}}\;m\).

Step-by-step solution

The term force may be defined as the product of mass and acceleration.

The term force may be defined as the product of mass and acceleration.

The term force may be defined as the product of mass and acceleration.

The mass of the\({\omega ^0}\)particle is\(783\;{\rm{MeV/}}{{\rm{c}}^{\rm{2}}}\), and the minimum uncertainty energy is:

\(\begin{array}{l}\Delta E = 783\;{\rm{MeV}}\left( {1.60 \times {{10}^{ - 13}}\;{\rm{J/MeV}}} \right)\\\Delta E = 1.25 \times {10^{ - 10}}\;{\rm{J}}\end{array}\)

The lifetime\(\delta t\)of the particle, according to the Heisenberg uncertainty principle:\(\begin{array}{l}\delta t = \frac{h}{{2\Delta E}}\\\delta t = \frac{{1.055 \times {{10}^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}}}{{2\left( {1.25 \times {{10}^{ - 10}}\;{\rm{J}}} \right)}}\\\delta t = 4.21 \times {10^{ - 25}}\;{\rm{s}}\end{array}\)

If we consider the speed of light is the average speed of the particle than

\(\begin{array}{l}r = c\delta t\\r = \left( {3 \times {{10}^8}\;{\rm{m/s}}} \right)\left( {4.21 \times {{10}^{ - 25}}\;{\rm{s}}} \right)\\r = 1.26 \times {10^{ - 16}}\;{\rm{m}}\end{array}\)

Hence, the range of the force is \(1.26 \times 1{0^{ - 16}}\;m\).