[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 N_y(t) = {exp(-_yt) - exp(-_xt)}. Find the time at which N_y is maximum and determine the population of x and Z at that instant.

Solution :-



also

Now





here the result





for y_{max      x}x = _yy      in that case only = 0





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

N_x = N₀e^{-1.5 ln 3} = = 1.92 x 10¹⁹

Also N_x + N_y + N_z = 10²⁰

=> N_z = 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.