Answer :

From the given diagram γ_{1} 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 γ_{1} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Thus, frequency of radiation radiated by γ_{2} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Thus, frequency of radiation radiated by γ_{3} decay is given by,

(∵ 1 eV = 1.6×10^{-19} C )

Given,

Mass of , m_{1} = 2.014102 u

Mass of , m_{2} = 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×c^{2}

= 0.001473 u×931.5 MeV/c^{2}

= 1.3720995 MeV

Since, β_{1} decays from maximum level to 1.088 MeV level then,

Kinetic energy of the β_{1} particle = (1.3720995 – 1.088) MeV

= 0.2840995 MeV

Since, β_{2} decays from maximum level to 0.412 MeV level then,

Kinetic energy of the β_{2} particle = (1.3720995 – 0.412) MeV

= 0.9600995 MeV

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