Question

Consider an ultrahigh-energy cosmic ray entering the Earth’s atmosphere (some have energies approaching a joule). Construct a problem in which you calculate the energy of the particle based on the number of particles in an observed cosmic ray shower. Among the things to consider are the average mass of the shower particles, the average number per square meter, and the extent (number of square meters covered) of the shower. Express the energy in \({\rm{eV}}\) and joules.

Short answer

The energy of the particle is obtained as \({\rm{1}}{{\rm{0}}^{{\rm{12}}}}{\rm{ TeV}}\).

Step-by-step solution

Concept Introduction

Incident energy is equivalent to the product of number of particles per square meter, energy of single particles and area on which particles are incident.

Incident energy = Number of particles per square meter\({\rm{ \times }}\)(energy of single particle\({\rm{ \times }}\)area on which particles are incident)

Information Provided

• Average mass of particles is: \({\rm{300 MeV/}}{{\rm{c}}^{\rm{2}}}\).
• Number of particles per square meter is: \({\rm{6}}{\rm{.67 \times 1}}{{\rm{0}}^{\rm{9}}}\).
• Energy of single particle is: \({\rm{0}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ eV}}\).
• Formula used is: Number of particles per square meter \({\rm{ = }}\) Incident energy \({\rm{/}}\) (energy of single particle \({\rm{ \times }}\) area on which particles are incident)

Calculation for Energy

The incident cosmic ray energy is calculated as –

Number of particles per square meter\({\rm{ = }}\)Incident energy\({\rm{/}}\)(energy of single particle\({\rm{ \times }}\)area on which particles are incident)

Incident energy = Number of particles per square meter\({\rm{ \times }}\)(energy of single particle\({\rm{ \times }}\)area on which particles are incident)

\(\begin{aligned}{}{E_i} &= {\rm{6}}{\rm{.67 \times 1}}{{\rm{0}}^{\rm{9}}} \times {\rm{0}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{6}}} \times {\rm{300 \times 1}}{{\rm{0}}^{\rm{6}}}\\ &= {\rm{1}}{{\rm{0}}^{{\rm{12}}}}{\rm{ TeV}}\end{aligned}\)

Therefore, the value for incident energy is obtained as \({\rm{1}}{{\rm{0}}^{{\rm{12}}}}{\rm{ TeV}}\).