Answer :
Answer:
2.42631E-13m
Explanation:
First we find the mass defect
Which is m= 0-2(9.10939E-33kg)
= - 1.82188E-30kg
Now find the energy
S
E= mc²=( -18.82188E-30)(2.999792E8)²
= 1.63742E-13J
Thus energy per photon will be
1.63742E-13J/2= 8.18710E-14J
So wavelength is given as
Lambda= hc/E
= (6.62608E-34)(2.997E8)/8.18710J
= 2.42631E-13m
The wavelength of radiation used to annihilate a positron and an electron is required.
The wavelength of the electromagnetic radiation used is 2.42 pm.
The mass of positron and electron are equal
m = Mass = [tex]\dfrac{0.000549}{6.022\times 10^{26}}=9.11\times 10^{-31}\ \text{kg}[/tex]
c = Speed of light = [tex]3\times 10^{8}\ \text{m/s}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]
Total energy in the collision is
[tex]E_T=mc^2+mc_2\\\Rightarrow E_T=2mc^2\\\Rightarrow E_T=2\times 9.11\times 10^{-31}\times (3\times 10^{8})^2\ \text{J}[/tex]
Energy released per photon is
[tex]E=\dfrac{E_T}{2}\\\Rightarrow E=\dfrac{2\times 9.11\times 10^{-31}\times (3\times 10^{8})^2}{2}\\\Rightarrow E=9.11\times 10^{-31}\times (3\times 10^{8})^2\ \text{J}[/tex]
Energy is given by
[tex]E=\dfrac{hc}{\lambda}\\\Rightarrow \lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{9.11\times 10^{-31}\times (3\times 10^{8})^2}\\\Rightarrow \lambda=2.42\times 10^{-12}\ \text{m}=2.42\ \text{pm}[/tex]
The wavelength of the electromagnetic radiation used is 2.42 pm.
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