Question

A newly discovered particle, the SPARTYON, has a mass 455 times that of an electron. If a SPARTYON at rest absorbs an anti-SPARTYON, what is the frequency of each of the emitted photons (in 1020 Hz)? The mass of an electron is 9.11×10-31 kg.

Answer 1

υ = 5.62 x 10¹⁴ Hz

Step-by-step explanation:

The process of of formation of photons takes place due to interaction of a particle with its anti-particle is called Annihilation of Matter. Here SPARTYON absorbs anti-SPARYTON to form 2 photons. Applying conservation of energy to the to this scenario, we get:

2(Energy of Each Photon) = Rest Mass Energy of SPARTYON + Rest Mass Energy of anti-SPARTYON

`2hv = m_(0)c^2 + m_(0)c^2\\\\2hv = 2m_(0)c^2\\\\hv = m_(0)c^2\\\\v = (m_(0)c^2)/(h)`

where,

υ = frequency of each photon = ?

m₀ = rest mass of SPARTYON = mass of anti-SPARTYON = 455*Mass of electron

m₀ = 455(9.1 x 10⁻³¹ kg) = 4.14 x 10⁻²⁸ kg

c = speed of light = 3 x 10⁸ m/s

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

Therefore, using these values in the equation, we get:

`v = ((4.14 x 10^(-28) kg)(3 x 10^8 m/s)^2)/(6.625 x 10^(-34)J.s)\\\\`

υ = 5.62 x 10¹⁴ Hz