Question

Another component of the strong nuclear force is transmitted by the exchange of virtual \[{\rm{K - }}\]mesons. Taking \[{\rm{K - }}\]-mesons to have an average mass of \[{\rm{495}}\;{\rm{MeV/}}{{\rm{c}}^{\rm{2}}}\], what is the approximate range of this component of the strong force?

Short answer

The approximate range of this component of string force is \[{\rm{1}}{\rm{.99 \times 1}}{{\rm{0}}^{{\rm{ - 16}}}}{\rm{\;m}}\].

Step-by-step solution

Definition of Heisenberg’s uncertainty principle

According to Heisenberg's uncertainty principle, it is impossible to precisely measure or compute an object's location and momentum simultaneously. The wave-particle duality of matter underpins this principle.

Step 2:Given Data

Average mass of K-mesons-\(m = 495\;{{{\rm{MeV}}} \mathord{\left/

{\vphantom {{{\rm{MeV}}} {{{\rm{c}}^{\rm{2}}}}}} \right.

\kern-\nulldelimiterspace} {{{\rm{c}}^{\rm{2}}}}}\)

Finding the approximate range of component

Using the uncertainty equation as a guide, we have-

\[\Delta E \cdot \Delta t = \frac{h}{{4\pi }}\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\]

Here,\[\Delta E\]is the uncertainty in energy and\[\Delta t\]is the uncertainty in time, and\[h\]is Plank’s constant.

we can get the range of this component,\[{\rm{d}}\], using the relation

\[\Delta t = \frac{d}{c} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)\]

Here, c is the speed of light in vacuum.

The uncertainty in energy can be calculated as-

\[\Delta E = m\left( {eV/{c^2}} \right)\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right)\]

Using equation (1), (2) and (3), we get-

\[\begin{array}{c}d = \frac{{hc}}{{4\pi \Delta E}}\\ = \frac{{\left( {4.14 \times {{10}^{ - 24}}\;{\rm{GeV}} \cdot {\rm{s}}} \right) \times \left( {3 \times {{10}^8}\;\;{\rm{m/s}}} \right)}}{{4\pi \times \left( {0.495\;{\rm{GeV}}} \right)}}\\ = 1.99 \times {10^{ - 16}}\;\;{\rm{m}}\end{array}\]

Therefore, the approximate range of this component is \[{\rm{1}}{\rm{.99 \times 1}}{{\rm{0}}^{{\rm{ - 16}}}}{\rm{\;m}}\].