An electron moving at 1000 m/s annihilates with a stationary positron. What is the frequency of the single photon produced?

The first thing to be aware of is that the frequency of a photon will depend on two things, its wavelength (c=fλ ) and its energy (E=hf). There is no information regarding the wavelength of the photon, but we can work out its energy using conservation of energy arguments, i.e. the total energy of the electron and positron will be equal to the energy of the photon because after the annihilation both the positron and electron will have disappeared.The total energy of the electron and positron will be the sum of their kinetic energies (KE=1/2mv2) and the energy associated with their masses (E=mc2). Only the electron is moving hence only the electron has kinetic energy which can be calculated by KEe = 1/2me10002 = 4.55510-25 J where me = 9.1110-31kg. The energy associated with the mass of the electron is Ee = mec2 = 8.19910-14 J. Because the positron is the antiparticle of the electron it has opposite charge but the same mass, hence its rest mass is also just Ee. Therefore, the total energy before the annihilation is 2Ee+ KEe = 1.639810-13 J. Using Ephoton= hf, we can divide our calculated value by the Planck constant (h = 6.6310-34 Js) to obtain the frequency, f, getting out f = 2.471020 Hz.

LM
Answered by Lawrence M. Physics tutor

1751 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

How can the first order kinematic (SUVAT) equations be derived?


What is meant by an excited atom?


A 10m long uniform beam is pivoted in its centre. A 30kg point mass is placed on one end of the beam. Where must a 50kg mass be placed in order to balance the beam?


A projectile is fired out of a cannon at 50 km/s, at an angle of 30 degrees and an elevation of 10m from the ground. How long does it take for the projectile to hit the ground?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences