The activity of a sample reduces from $A_0$ to $A_0 / \sqrt{3}$ in one hour. The activity after $3$ hours more will be

  • A
    $\frac{A_0}{3\sqrt{3}}$
  • B
    $\frac{A_0}{9}$
  • C
    $\frac{A_0}{9\sqrt{3}}$
  • D
    $\frac{A_0}{27}$

Explore More

Similar Questions

The activity of a radioactive sample falls from $700 \; s^{-1}$ to $500 \; s^{-1}$ in $30 \; min$. Its half-life is close to ......... $min$.

The number of nuclei of a radioactive substance is $1000$ and $900$ at times $t = 0$ and $t = 2 \ s$,respectively. Then,the number of nuclei at time $t = 4 \ s$ will be:

Difficult
View Solution

$A$ radioactive sample has a half-life of $5$ years. The percentage of the fraction decayed in $10$ years will be (in $\%$)

The activity of a radioactive substance is $R_1$ at time $t_1$ and $R_2$ at time $t_2$. If $\lambda$ is the decay constant,then which of the following is correct?

If the measurement errors in all the independent quantities are known,then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example,consider the relation $z = x / y$. If the errors in $x, y$ and $z$ are $\Delta x, \Delta y$ and $\Delta z$,respectively,then $z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y} (1 \pm \frac{\Delta x}{x}) (1 \pm \frac{\Delta y}{y})^{-1}$. The series expansion for $(1 \pm \frac{\Delta y}{y})^{-1}$,to first power in $\Delta y / y$,is $1 \mp (\Delta y / y)$. The relative errors in independent variables are always added. So the error in $z$ will be $\Delta z = z (\frac{\Delta x}{x} + \frac{\Delta y}{y})$. The above derivation makes the assumption that $\Delta x / x \ll 1, \Delta y / y \ll 1$. Therefore,the higher powers of these quantities are neglected.
$(1)$ Consider the ratio $r = \frac{(1 - a)}{(1 + a)}$ to be determined by measuring a dimensionless quantity $a$. If the error in the measurement of $a$ is $\Delta a$ $(\Delta a / a \ll 1)$,then what is the error $\Delta r$?
$(2)$ In an experiment,the initial number of radioactive nuclei is $3000$. It is found that $1000 \pm 40$ nuclei decayed in the first $1.0 \ s$. For $|x| < 1$,$\ln(1 + x) = x$ up to first power in $x$. The error $\Delta \lambda$,in the determination of the decay constant $\lambda$,in $s^{-1}$,is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo