A sample initially contains only U-238 isotop | vedclass

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

Question

(A) Let the initial amount of $U-238$ be $N_0$ atoms. At time $t$,let $N_t$ be the number of $U-238$ atoms remaining and $N_{Pb}$ be the number of $Pb

Vedclass · Accepted Answer

$143$

Vedclass · Answer

(A) Let the initial amount of $U-238$ be $N_0$ atoms. At time $t$,let $N_t$ be the number of $U-238$ atoms remaining and $N_{Pb}$ be the number of $Pb-206$ atoms formed.Since $1$ atom of $U-238$ produces $1$ atom of $Pb-206$,$N_0 = N_t + N_{Pb}$.The mass ratio is given as $\frac{m_{Pb}}{m_U} = 7$. Since $m = \frac{N 	imes M}{N_A}$,we have $\frac{N_{Pb} 	imes 206}{N_t 	imes 238} = 7$.Thus,$N_{Pb} = 7 	imes N_t 	imes \frac{238}{206} = N_t 	imes \frac{1666}{206} \approx 8.087 N_t$.Then $N_0 = N_t + 8.087 N_t = 9.087 N_t$.Using the radioactive decay law: $N_t = N_0 e^{-\lambda t}$,so $\frac{N_0}{N_t} = e^{\lambda t}$.$\lambda t = \ln(9.087) \approx 2.2068$.Given $T_{1/2} = 4.5 	imes 10^9$ years,$\lambda = \frac{\ln 2}{4.5 	imes 10^9} \approx \frac{0.693}{4.5 	imes 10^9}$.$t = \frac{2.2068 	imes 4.5 	imes 10^9}{0.693} \approx 14.33 	imes 10^9 = 143.3 	imes 10^8$ years.Thus,$P \approx 143$.

Similar Questions

$A$ radioactive element which can decay by two processes has half-life $t_1$ for the first process and half-life $t_2$ for the second process. Let $\langle t \rangle$ be the effective average-life of this element. Which of the following is correct?

The half-life of a radioactive substance is $12 \text{ minutes}$. The time gap between $28 \%$ decay and $82 \%$ decay of the radioactive substance is

An element $X$ with a half-life of $1.4 \times 10^9$ years decays to form another stable element $Y$. $A$ sample is taken from a rock that contains both $X$ and $Y$ in the ratio $1:7$. If at the time of formation of the rock,$Y$ was not present in the sample,then the age of the rock in years is

The half-life of a sample of a radioactive substance is $1 \text{ hour}$. If $8 \times 10^{10}$ atoms are present at $t = 0$,then the number of atoms decayed in the duration $t = 2 \text{ hours}$ to $t = 4 \text{ hours}$ will be

The rate of disintegration was observed to be $10^{17}$ disintegrations per second when its half-life period is $1445$ years. The original number of particles is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module