The half-life of a radioactive isotope X is 5 | vedclass

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

Question

(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount of $X$.The amount of stable element $Y$ formed is $N_0

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount of $X$.The amount of stable element $Y$ formed is $N_0 - N$.Given the ratio of $X$ to $Y$ is $1:15$,we have $\frac{N}{N_0 - N} = \frac{1}{15}$.Adding $1$ to both sides or rearranging gives $\frac{N_0 - N}{N} = 15$,so $\frac{N_0}{N} - 1 = 15$,which implies $\frac{N_0}{N} = 16$.We know the radioactive decay law is $\frac{N}{N_0} = (\frac{1}{2})^{t/T_{1/2}}$,where $T_{1/2} = 50$ years.Thus,$\frac{N_0}{N} = 2^{t/T_{1/2}} = 16$.Since $16 = 2^4$,we have $\frac{t}{T_{1/2}} = 4$.Therefore,$t = 4 	imes 50 = 200$ years.

The half-life of a radioactive isotope $X$ is $50$ years. It decays to another element $Y$,which is stable. The two elements $X$ and $Y$ were found to be in the ratio of $1:15$ in a sample of a given rock. The age of the rock was estimated to be ......... years.

Similar Questions

$A$ particle is performing simple harmonic motion along the $x$-axis with an amplitude of $4\, cm$ and a time period of $1.2\, s$. The minimum time taken by the particle to move from $x = 2\, cm$ to $x = +4\, cm$ and back again is given by .... $s$.

$A$ graph of magnetic flux $(\phi)$ versus current $(I)$ is drawn for four inductors $A, B, C, D$. Which inductor has the largest value of self-inductance?

$\mathop {\lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = $

Which of the following statements is correct?

Which of the following is commonly used in a desiccator in the laboratory?

Vedclass Test Series

Exam Paper Generator

Online Exam Module