Q. 295.0( 2 Votes )

Obtain the maximu

Answer :

From the given diagram γ1 decays from the 1.088 MeV energy level to 0 MeV level.


We have,


E = hν


Where,


h = plank’s constant = 6.6×10-34 J s


ν = frequency


Thus, frequency of radiation radiated by γ1 decay is given by,



( 1 eV = 1.6×10-19 C )


Thus, frequency of radiation radiated by γ2 decay is given by,



( 1 eV = 1.6×10-19 C )


Thus, frequency of radiation radiated by γ3 decay is given by,



( 1 eV = 1.6×10-19 C )


Given,


Mass of , m1 = 2.014102 u


Mass of , m2 = 3.016049 u


The energy of the highest level is given by,



Where,


Δm = Mass defect (or) mass lost during reaction


c = speed of light


E = (197.968233 – 197.966760) u×c2


= 0.001473 u×931.5 MeV/c2


= 1.3720995 MeV


Since, β1 decays from maximum level to 1.088 MeV level then,


Kinetic energy of the β1 particle = (1.3720995 – 1.088) MeV


= 0.2840995 MeV


Since, β2 decays from maximum level to 0.412 MeV level then,


Kinetic energy of the β2 particle = (1.3720995 – 0.412) MeV


= 0.9600995 MeV


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