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Nucleus
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Nucleus


[2] A radioactive nucleus x decays to a nucleus y with a decay constant x = 0.1 s-1

y further decays to a stable nucleus z with a decay constant y = 1/30. Intially, there are only x nuclei and their number is N0 = 1020. Set up the rate equations for the populations of x, y, and z. The population of the y nucleus as a function of time is given by Ny(t) = {exp(-yt) - exp(-xt)}. Find the time at which Ny is maximum and determine the population of x and Z at that instant.

Solution :-



also

Now





here the result





for ymax      xx = yy      in that case only = 0





tmax = 15 ln = 15 ln 3 = 15 x 2.303 x log3 = 16.47 s

Nx = N0e-1.5 ln 3 = = 1.92 x 1019

Also Nx + Ny + Nz = 1020

=> Nz = 1020 - 1.92 x 1019 - 5.7 x 1019

           = 2.32 x 1019
Keywords:

° Nucleus

° Mass defect

° Binding energy

° Nuclear Fusion

° Rdioactivity

° Rate constant

° Half life

° Mean/Average life.
 

 
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