Two radioactive nuclei P and Q in a given sam | vedclass

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(D) At $t = 0,$ $N_P(0) = 4N_0$ and $N_Q(0) = N_0.$The number of nuclei at time $t$ is given by $N(t) = N(0) \cdot (1/2)^{t/T_{1/2}}.$For $P,$ $N_P(t)

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$\frac{9N_0}{2}$

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(D) At $t = 0,$ $N_P(0) = 4N_0$ and $N_Q(0) = N_0.$The number of nuclei at time $t$ is given by $N(t) = N(0) \cdot (1/2)^{t/T_{1/2}}.$For $P,$ $N_P(t) = 4N_0 \cdot (1/2)^{t/1} = 4N_0 \cdot 2^{-t}.$For $Q,$ $N_Q(t) = N_0 \cdot (1/2)^{t/2} = N_0 \cdot 2^{-t/2}.$Given $N_P(t) = N_Q(t),$ we have $4N_0 \cdot 2^{-t} = N_0 \cdot 2^{-t/2}.$Dividing by $N_0,$ we get $4 = 2^t / 2^{t/2} = 2^{t/2}.$Since $4 = 2^2,$ we have $t/2 = 2,$ so $t = 4 	ext{ minutes}.$At $t = 4 	ext{ minutes},$$N_P(4) = 4N_0 \cdot (1/2)^4 = 4N_0 / 16 = N_0/4.$$N_Q(4) = N_0 \cdot (1/2)^{4/2} = N_0 / 4.$The number of nuclei of $R$ formed is the total number of decayed nuclei of $P$ and $Q.$$N_R = (N_P(0) - N_P(4)) + (N_Q(0) - N_Q(4)).$$N_R = (4N_0 - N_0/4) + (N_0 - N_0/4) = 15N_0/4 + 3N_0/4 = 18N_0/4 = 9N_0/2.$

$t (h)$	$0$	$1$	$2$	$3$	$4$
$R (MBq)$	$100$	$35.36$	$12.51$	$4.42$	$1.56$

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