Question

Plans for an accelerator that produces a secondary beam of \({\rm{K}}\)-mesons to scatter from nuclei, for the purpose of studying the strong force, call for them to have a kinetic energy of \({\rm{500 MeV}}\).

(a) What would the relativistic quantity \(\gamma {\rm{ = }}\frac{{\rm{1}}}{{\sqrt {{\rm{1 - }}{{{{\rm{\nu }}^{\rm{2}}}} \mathord{\left/{\vphantom {{{{\rm{\nu }}^{\rm{2}}}} {{{\rm{c}}^{\rm{2}}}}}} \right. \\} {{{\rm{c}}^{\rm{2}}}}}} }}\) be for these particles?

(b) How long would their average lifetime be in the laboratory?

(c) How far could they travel in this time?

Short answer

(a) The relativistic quantity\(\gamma = \frac{1}{{\sqrt {1 - {v^2}/{c^2}} }}\)for the particles is\(\gamma = 2.013\).

(b) The average lifetime of the particles in the laboratory will be\(\Delta t = 2.50 \times {10^{ - 8}}\;{\rm{s}}\).

(c) The particles could travel a distance of \(d = 6.51{\rm{\;m}}\) in the time \(\Delta t = 2.50 \times {10^{ - 8}}\;{\rm{s}}\).

Step-by-step solution

Concept Introduction

The relation between mass and energy is given by the energy expression,

\(E = \Delta m{c^2}\)

Here\(E\)is the energy of the physical system,\(m\)is the mass of the system and\(c\)is the speed of the light in vacuum.

Calculation for relativistic quantity

(a)

Kinetic energy can be calculated as,

\({E_{kin{\rm{ }}}} = (\gamma - 1)m{c^2}\)……. (i)

Where\(m\)is the kaon’s rest mass. As one is creating an accelerator beam, the particles needs to be charged to be accelerated, so consider a\({K^ + }\)or\({K^ - }\)meson (this is relevant because the mass of\({K^ \pm }\)differs from\({K^0}\)). Inverting\({\rm{(1)}}\)to express\(\gamma \)and inputting the known values, it is obtained as,

\(\begin{aligned}{}\gamma &= \frac{{{E_{kin{\rm{ }}}}}}{{m{c^2}}} + 1\\ &= \frac{{{\rm{(500 MeV)}}}}{{\left( {{\rm{493}}{\rm{.67 MeV/}}{{\rm{c}}^{\rm{2}}}} \right){{\rm{c}}^{\rm{2}}}}}{\rm{ + 1}}\\ &= {\rm{2}}{\rm{.013}}\end{aligned}\)

Therefore, the value for relativistic quantity is obtained as \(\gamma = 2.013\).

Lifetime of the particle

(b)

The proper lifetime of the kaon’s is \(\Delta {t_0} = 1.24 \times {10^{ - 8}}{\rm{\;s}}\) so to calculate its average lifetime in the laboratory \({\rm{\Delta t}}\)

\(\begin{aligned}{}\Delta t &= \gamma \Delta {t_0}\\ &= (2.013) \cdot \left( {1.24 \times {{10}^{ - 8}}\;{\rm{s}}} \right)\\ &= 2.50 \times {10^{ - 8}}{\rm{\;s}}\end{aligned}\)

Therefore, the value for time is obtained as \({\rm{\Delta t = 2}}{\rm{.50 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{\;s}}\).

Calculation for distance

(c)

The equation for the relativistic quantity is given by,

\(\gamma = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)

And invert this equation to get –

\(v = c\sqrt {1 - \frac{1}{{{\gamma ^2}}}} \)

The distance travelled\(d\)is given by,

\(\begin{aligned}{}d &= v\Delta t\\ &= c\sqrt {1 - \frac{1}{{{\gamma ^2}}}} \Delta t\\ &= \left( {{\rm{3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m}}{{\rm{s}}^{{\rm{ - 1}}}}} \right)\sqrt {{\rm{1 - }}\frac{{\rm{1}}}{{{\rm{2}}{\rm{.01}}{{\rm{3}}^{\rm{2}}}}}} \cdot \left( {{\rm{2}}{\rm{.50 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{\;s}}} \right)\\ &= 6.51\;{\rm{m}}\end{aligned}\)

Therefore, the value for distance is obtained as \(d = 6.51\;{\rm{m}}\).