Starting with a sample of pure ^{66}Cu,frac{7 | vedclass

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

Question

(A) Given that $\frac{7}{8}$ of the sample decays,the remaining fraction of the sample is $1 - \frac{7}{8} = \frac{1}{8}$.The law of radioactive decay

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given that $\frac{7}{8}$ of the sample decays,the remaining fraction of the sample is $1 - \frac{7}{8} = \frac{1}{8}$.The law of radioactive decay is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.Substituting the values: $\frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{15}{T_{1/2}}}$.Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$,we have $\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{\frac{15}{T_{1/2}}}$.Equating the exponents: $3 = \frac{15}{T_{1/2}}$.Therefore,$T_{1/2} = \frac{15}{3} = 5 \ min$.

Starting with a sample of pure $^{66}Cu$,$\frac{7}{8}$ of it decays into $Zn$ in $15 \ min$. The corresponding half-life is .......... $min$.

Similar Questions

The differential equation of all circles which pass through the origin and whose centres lie on the $Y$-axis is

Among the species given below, the total number of diamagnetic species is. . . . . $H$ atom, $NO _2$ monomer, $O _2^{-}$(superoxide), dimeric sulphur in vapour phase, $MnsO _4$, $\left( NH _4\right)_2\left[ FeCl _4\right],\left( NH _4\right)_2\left[ NiCl _4\right], K _2 MnO _4, K _2 CrO _4$

If $I_n = \int \sin^n x \, dx$,then $n I_n - (n-1) I_{n-2}$ equals

The figure shows a system of logic gates. From the study of the truth table,it can be found that to produce a high output $(1)$ at $R$,we must have:

Vedclass Test Series

Exam Paper Generator

Online Exam Module