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Number of P species present at time t=0 is 4No
Half life of P is 1 minute
After 1st half life the number of P species = 2No
After 2nd half life the number of P species = No
After 3rd half life the number of P species = No/2
After 4th half life the number of P species = No/4
Total time = 4 minutes.
Hence, R = 4No– No/4
Number of Q species present at time t=0 is No
Half life of Q is 2 minutes
After 1st half life the number of Q species = No/2
After 2nd half life the number of Q species = No/4
Total time = 4 minutes
Hence, R = No – No/4
Total R = [4No – No/4] + No – No/4 = 15No/4 +3/4 No = 18No/4 = 9No/2
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