A gamma ray photon creates an electron-positr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The process of pair production involves a gamma ray photon interacting with a nucleus to produce an electron and a positron.The energy of the gamm

Vedclass · Accepted Answer

$1.78$

Vedclass · Answer

(B) The process of pair production involves a gamma ray photon interacting with a nucleus to produce an electron and a positron.The energy of the gamma ray photon $(E_{\gamma})$ is used to provide the rest mass energy of the electron $(m_e c^2)$,the rest mass energy of the positron $(m_p c^2)$,and the total kinetic energy $(K.E.)$ of the pair.Given:Rest mass energy of electron $(m_e c^2)$ = $0.5\, MeV$.Rest mass energy of positron $(m_p c^2)$ = $0.5\, MeV$ (since it is the antiparticle of the electron).Total kinetic energy $(K.E.)$ = $0.78\, MeV$.Therefore,the energy of the gamma ray photon is:$E_{\gamma} = m_e c^2 + m_p c^2 + K.E.$$E_{\gamma} = 0.5\, MeV + 0.5\, MeV + 0.78\, MeV = 1.78\, MeV$.

$A$ gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is $0.5\, MeV$ and the total kinetic energy $(K.E.)$ of the electron-positron pair is $0.78\, MeV$,then the energy of the gamma ray photon must be.......$MeV$.

Similar Questions

Can a quantum of light be associated with a particle?

Monochromatic light of wavelength $3000 \, \mathring{A}$ is incident on a surface of area $4 \, cm^2$. If the intensity of light is $150 \, mW/m^2$, then the rate at which photons strike the target is:

Write the equation for the energy of a photon.

Let $n_r$ and $n_b$ be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.

$A$ radio transmitter radiates $1 \text{ kW}$ power at a wavelength of $198.6 \text{ m}$. How many photons does it emit per second?

Vedclass Test Series

Exam Paper Generator

Online Exam Module