Two radioactive isotopes P and Q have half-lives of 10 min and 15 min , respectively. Initially, the number of nuclei of P is 4 times the number of nuclei of Q . The time after which both isotopes will have the same number of nuclei remaining is

Step-by-step solution

Let the initial number of nuclei of isotope Q be N₀ . Then, the initial number of nuclei of isotope P is 4N₀ . Using the radioactive decay law, the number of nuclei remaining after time t is N = N₀ ( 1 2 )^ t T_ 1/2 . For isotope P : N_P = 4N₀ ( 1 2 )^ t 10 For isotope Q : N_Q = N₀ ( 1 2 )^ t 15 We are given that N_P = N_Q at time t : 4N₀ ( 1 2 )^ t 10 = N₀ ( 1 2 )^ t 15 Cancelling N₀ and rearranging: 4 = ( 1 2 )^ t 15 ( 1 2 )^ t 10 4 = ( 1 2 )^ t 15 - t 10 4 = ( 1 2 )^ 2t - 3t 30 = ( 1 2 )^ - t 30 = 2^ t 30 Since