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Debora [2.8K]
3 years ago
10

A positron and an electron annihilate each other upon colliding, thereby producing energy in the form of two gamma rays. Assumin

g that both gamma rays have the same energy (since both particles have the same mass), calculate the wavelength of the electromagnetic radiation used in pm. (1 pm = 10⁻¹² m; mass of electron (amu) = 0.000549)
Physics
1 answer:
kolbaska11 [484]3 years ago
6 0

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

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