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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $N_P(t)$ and $N_Q(t)$ be the number of nuclei of $P$ and $Q$ at time $t$. Given: $N_P(0) = 4N_0$,$T_{1/2, P} = 1$ min,$N_Q(0) = N_0$,$T_{1/2,

Vedclass · Accepted Answer

$\frac{9N_0}{2}$

Vedclass · Answer

(C) Let $N_P(t)$ and $N_Q(t)$ be the number of nuclei of $P$ and $Q$ at time $t$. Given: $N_P(0) = 4N_0$,$T_{1/2, P} = 1$ min,$N_Q(0) = N_0$,$T_{1/2, Q} = 2$ min. The decay equations are: $N_P(t) = 4N_0 \cdot (1/2)^{t/1}$ and $N_Q(t) = N_0 \cdot (1/2)^{t/2}$. We need to find $t$ when $N_P(t) = N_Q(t)$: $4N_0 \cdot (1/2)^t = N_0 \cdot (1/2)^{t/2}$ $4 = (1/2)^{t/2} / (1/2)^t = (1/2)^{-t/2} = 2^{t/2}$ $2^2 = 2^{t/2} \implies t/2 = 2 \implies t = 4$ minutes. At $t = 4$ min: $N_P(4) = 4N_0 \cdot (1/2)^4 = 4N_0 / 16 = N_0/4$. $N_Q(4) = N_0 \cdot (1/2)^{4/2} = N_0 / 4$. The number of nuclei of $R$ formed is the total number of decayed nuclei of $P$ and $Q$: $N_R = (N_P(0) - N_P(4)) + (N_Q(0) - N_Q(4))$ $N_R = (4N_0 - N_0/4) + (N_0 - N_0/4) = 15N_0/4 + 3N_0/4 = 18N_0/4 = 9N_0/2$.

Similar Questions

What is $X$ in the following reaction?
$2 CH_3COCH_3 \stackrel{Ba(OH)_2}{\longrightarrow} X$

The product of the lengths of the perpendiculars drawn from the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ to the tangent at any point on the ellipse is

The potential energy of a particle oscillating along the $x-$axis is given as $U = 20 + (x - 2)^2$,where $U$ is in $joules$ and $x$ is in $meters$. The total mechanical energy of the particle is $36 \, J$. The maximum kinetic energy of the particle is ............... $J$.

One mole of alkene $\underline{X}$ on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The $IUPAC$ name of $\underline{X}$ is

The major product in the following reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module