A certain radioactive substance reduces to 25 | vedclass

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

Question

(B) The radioactive decay law is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}$,where $T$ is the half-life.Given that the substance reduces to $

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The radioactive decay law is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}$,where $T$ is the half-life.Given that the substance reduces to $25\%$ of its initial value,we have $N(t) = 0.25 N_0 = \frac{1}{4} N_0$.Substituting this into the equation: $\frac{1}{4} N_0 = N_0 \left(\frac{1}{2}\right)^{16/T}$.This simplifies to $\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{16/T}$.Equating the exponents,we get $2 = \frac{16}{T}$.Solving for $T$,we find $T = \frac{16}{2} = 8\, days$.

$A$ certain radioactive substance reduces to $25\%$ of its initial value in $16\ days$. Its half-life is .......... $days$.

Similar Questions

$A$ radioactive material of half-life $2.5$ hours emits radiation that is $32$ times the safe maximum level. The time (in hours) after which the material can be handled safely is

$A$ radioactive substance emits $100$ beta particles in the first $2 \,s$ and $50$ beta particles in the next $2 \,s$. The mean life of the sample is

The half-life period of a radioactive substance is $5\, min$. The amount of substance decayed in $20\, min$ will be..........$\%$

In a radioactive disintegration,the ratio of the initial number of atoms to the number of atoms present at an instant of time equal to its mean life is:

If the half-life of a radioactive atom is $2.3 \, days$,then its decay constant would be

Vedclass Test Series

Exam Paper Generator

Online Exam Module