Question

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

Answer 1

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