Question

In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy $10.2BeV$ into two $\gamma $–rays of equal energy. What is the wavelength associated with each $\gamma $–rays?$(1BeV = {10^9}eV)$

Answer 1

Hint: The total energy of the electron-positron pair is given and it is equal to the energy of two $\gamma $–rays. So you can find the energy of one $\gamma$–ray. Must represent the energy in the Joule unit. To find the wavelength associated with each $\gamma $–ray, use the relation between the energy and wavelength of a particle.

Formula used:
The energy of each $\gamma $–ray, $E' = \dfrac{E}{2}$
Where $E$ =The total energy of electron-positron pair =energy of two $\gamma $–rays
$E$ = $\dfrac{{hc}}{\lambda }$
Where $h$= the Planck’s constant.
$c$= the speed of light,
$\lambda $= the wavelength associated with each $\gamma $–ray.

Complete step by step answer:
The high energy collision between the electron and positron in a high accelerator experiment results in an interpretation of an event as the annihilation of an electron-positron pair.
The total energy of the electron-positron pair = $E$
Given, $E = 10.2BeV$
Since, $1BeV = {10^9}eV$and $1eV = 1.6 \times {10^{ - 19}}J$
$\therefore E = 10.2 \times {10^9} \times 1.6 \times {10^{ - 19}}J$
On multiplying the terms and we get,
$ \Rightarrow E = 16.32 \times {10^{ - 10}}J$
This energy is equal to the energy of two $\gamma $–rays.
Hence, The energy of each $\gamma $–ray, $E' = \dfrac{E}{2}$
$ \Rightarrow E' = \dfrac{{16.32 \times {{10}^{ - 10}}}}{2}$
Let us divide the terms and we get
$ \Rightarrow E' = 8.16 \times {10^{ - 10}}J$
Now, we know the relation between the energy of a photon with its wavelength is, $E = \dfrac{{hc}}{\lambda }$
Where $h$= the Planck’s constant = $6.625 \times {10^{ - 34}}Js$
 $c$= the speed of light =$3 \times {10^8}m/s$
Hence, if the wavelength associated with each $\gamma $–rays is $\lambda $,
$\therefore \lambda = \dfrac{{hc}}{{E'}}$
$ \Rightarrow \lambda = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{8.16 \times {{10}^{ - 10}}}}$
On simplifying we get, $ \Rightarrow \lambda = 2.43 \times {10^{ - 16}}m$.
Additional information:
$\gamma $–ray is an electromagnetic wave like the visible light; the speed of $\gamma $–ray is equal to the speed of light in any medium. The wavelength of $\gamma $–ray is in the range of $1{A^0}$ to ${10^{ - 2}}{A^0}$. There is no effect of the electric field or magnetic field on the path of $\gamma $–ray.
Note: Here we use the energy- wavelength equation of a photon particle to calculate the wavelength associated with the gamma-ray and also put the value of the energy of each gamma-ray in the position of energy. This is because according to quantum theory $\gamma $–a ray is the flow of photon particles with very high energy. The energy can be up to a few Mega Electron Volts ($MeV$).