Nucleus - 7

[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.