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Question: In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event ...

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 10.2BeV10.2BeV into two γ\gamma –rays of equal energy. What is the wavelength associated with each γ\gamma –rays?(1BeV=109eV)(1BeV = {10^9}eV)

Explanation

Solution

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

Formula used:
The energy of each γ\gamma –ray, E′=E2E' = \dfrac{E}{2}
Where EE =The total energy of electron-positron pair =energy of two γ\gamma –rays
EE = hcλ\dfrac{{hc}}{\lambda }
Where hh= the Planck’s constant.
cc= the speed of light,
λ\lambda = the wavelength associated with each γ\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 = EE
Given, E=10.2BeVE = 10.2BeV
Since, 1BeV=109eV1BeV = {10^9}eVand 1eV=1.6×10−19J1eV = 1.6 \times {10^{ - 19}}J
∴E=10.2×109×1.6×10−19J\therefore E = 10.2 \times {10^9} \times 1.6 \times {10^{ - 19}}J
On multiplying the terms and we get,
⇒E=16.32×10−10J\Rightarrow E = 16.32 \times {10^{ - 10}}J
This energy is equal to the energy of two γ\gamma –rays.
Hence, The energy of each γ\gamma –ray, E′=E2E' = \dfrac{E}{2}
⇒E′=16.32×10−102\Rightarrow E' = \dfrac{{16.32 \times {{10}^{ - 10}}}}{2}
Let us divide the terms and we get
⇒E′=8.16×10−10J\Rightarrow E' = 8.16 \times {10^{ - 10}}J
Now, we know the relation between the energy of a photon with its wavelength is, E=hcλE = \dfrac{{hc}}{\lambda }
Where hh= the Planck’s constant = 6.625×10−34Js6.625 \times {10^{ - 34}}Js
cc= the speed of light =3×108m/s3 \times {10^8}m/s
Hence, if the wavelength associated with each γ\gamma –rays is λ\lambda ,
∴λ=hcE′\therefore \lambda = \dfrac{{hc}}{{E'}}
⇒λ=6.625×10−34×3×1088.16×10−10\Rightarrow \lambda = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{8.16 \times {{10}^{ - 10}}}}
On simplifying we get, ⇒λ=2.43×10−16m \Rightarrow \lambda = 2.43 \times {10^{ - 16}}m.
Additional information:
γ\gamma –ray is an electromagnetic wave like the visible light; the speed of γ\gamma –ray is equal to the speed of light in any medium. The wavelength of γ\gamma –ray is in the range of 1A01{A^0} to 10−2A0{10^{ - 2}}{A^0}. There is no effect of the electric field or magnetic field on the path of γ\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 γ\gamma –a ray is the flow of photon particles with very high energy. The energy can be up to a few Mega Electron Volts (MeVMeV).