Question

A gamma-ray photon produces an electron and a positron, each with a kinetic energy of 261 keV . h=6.626×10−34J⋅s, c=2.998×108m/s, me=9.109×10−31kg, e=1.602×10−19C.
A) Determine the energy of the photon.B) Determine the wavelength of the photon.

Answer 1

Answer:

The energy and the wavelength of the photon are 1.546 MeV and $8.036\times10^{-13}\ m$.

Explanation:

Given that,

Kinetic energy = 261 KeV

Planck's constant $h = 6.626\times10^{−34}\ J.s$

Speed of light $c=2.998\times10^{8}\ m/s$

Mass of electron $m_{e}=9.109\times10^{-31}\ kg$

Charge $q=1.602\times10^{-19}\ C$

(A). We need to calculate the energy of the photon

Using formula of rest mass energy

$E=m_{0}c^2$

$E=9.109\times10^{-31}\times(3\times10^{8})^2$

$E=8.198\times10^{-14}\ J$

Energy in eV

$E=\dfrac{8.198\times10^{-14}}{1.6\times10^{-19}}$

$E=512375\ eV$

$E=0.512\ MeV$

The total energy of photon

$TE=2(E+K.E)$

$TE=2(0.512+0.261)$

$TE=1.546\ MeV$

(B). We need to calculate the wavelength of the photon

Using formula of wavelength

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

$\lambda=\dfrac{6.626\times10^{−34}\times3\times10^{8}}{1.546\times10^{6}\times1.6\times10^{-19}}$

$\lambda=8.036\times10^{-13}\ m$

Hence, The energy and the wavelength of the photon are 1.546 MeV and $8.036\times10^{-13}\ m$.