# Obtain the maximu

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

( 1 eV = 1.6×10-19 C )

( 1 eV = 1.6×10-19 C )

( 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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