A radio isotope X with a half life 1.4 10^8 year, decays to Y, which is stable. A sample of rock from a cave was found to contain X and Y in the ratio 1 : 7 . The age of rock (in year) is

Step-by-step solution

Let the initial number of atoms of X be N₀ . Since X decays to Y, the number of atoms of Y formed is N_Y = N₀ - N_X , where N_X is the number of atoms of X remaining. Given the ratio of X to Y is 1 : 7 , we have N_X N_Y = 1 7 . Substituting N_Y = N₀ - N_X , we get N_X N₀ - N_X = 1 7 . 7N_X = N₀ - N_X 8N_X = N₀ N_X = N₀ 8 . The amount of X remaining can be written as N_X = N₀ ( 1 2 )^n , where n is the number of half-lives. N₀ 8 = N₀ ( 1 2 )^n ( 1 2 )^3 = ( 1 2 )^n n = 3 . The age of the rock is t = n T_ 1/2 . t = 3