Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t=0, the number of P species are 4N0 and that of Q are N0. Half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. Initially, there are no nuclei of R present in the sample. When the number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be?

Initially P → 4N0
Q → N0
Half life, TP = 1 min
Tq = 2 min
Let after time t number of nuclei of P and Q are equal that is:
4N0e^(-λPt) = N0e^(-λQt)
4e^(-λPt) = e^(-λQt)
4e^(-0.693t/1) = e^(-0.693t/2)
4e^(-0.693t) = e^(-0.3465t)
Taking logarithm on both sides:
ln4 + ln(e^(-0.693t)) = ln(e^(-0.3465t))
1.386 - 0.693t = -0.3465t
1.386 = 0.3465t
t = 4 min
Number of P remaining after 4 min = 4N0e^(-0.6934) = 4N0e^(-2.772) = 4N0(0.0625) = 0.25N0
Number of Q remaining after 4 min = N0e^(-0.6934/2) = N0e^(-1.386) = N0(0.25) = 0.25N0
Number of R = 4N0 - 0.25N0 + N0 - 0.25N0 = 4N0 - 0.5N0 = 3.5N0 ≈ 2N0
Another method:
Number of P remaining = 4N0(1/2)^t = 4N0(1/2)^4 = N0/4
Number of Q remaining = N0(1/2)^t/2 = N0(1/2)^2 = N0/4
Number of R = 4N0 - N0/4 + N0 - N0/4 = 9N0/2