# Nuclear Chemistry

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The half- life of a positron is very short. it reacts with an electron, and the masses of both are converted to two gamma-ray photons:
(0e+1)+ (0e-1)---> 2γ
(the 0 in front of e is the mass # and behind it is the atomic #)
(γ stands for the photons)
This reaction is called an annihilation reaction. the mass of an electrion or positron is 9.109 * 10^-31 kg
a) Calculate the energy produced by the reaction between one electron and one positron.( in Joules)
change in E= ______J
b)Assuming that the two γ-ray photons have the same frequency,calculate this frequency.
γ photon=_____Hz

• Nuclear Chemistry -

a) E = 2 m c^2,
where m is the electron (or positron) mass, 9.109 * 10^-31 kg
and c is the speed of light

b) h*f = E/2 = m c^2
Solve for frequency, f.

• Nuclear Chemistry -

in b) what does h equal

• Nuclear Chemistry -

h=6.626*10^-34 J*s

• Phsyical science -

What physical quantities are governed by the quantum numbers of an atomic electron?.

• Nuclear Chemistry -

?

• Nuclear Chemistry -

Part A Calculate the energy by the reaction between one electron and one positron.
E=(2)mc^2
E=(2)(9.109x10^-31 kg)(3.00x10^8)^2
E=1.63962x10^-13 J

Part B Assuming that the two gamma ray photons have the same frequency.
h*f= E/2= mc^2 solve for f
(6.626x10^-34 J*s)(f) =(1.63962x10^-13/ 2 ) = (9.109x10^-31)(3.00x10^8)^2
f=(8.1981x10^-14) / (6.626x10^-34) = 8.1981x10^-14
f=1.2372623x10^20 = 8.1981x10^-14
f=1.2372623x10^20 + -8.1981x10^-14
f=1.2372623x10^20 Hz

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