If 20 , g of a radioactive substance reduces | vedclass

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

Question

(B) The half-life $T_{1/2}$ is the time taken for a substance to reduce to half its initial mass.Given that $20 \, g$ reduces to $10 \, g$ in $4 \, mi

Vedclass · Accepted Answer

In $12 \, minutes$

Vedclass · Answer

(B) The half-life $T_{1/2}$ is the time taken for a substance to reduce to half its initial mass.Given that $20 \, g$ reduces to $10 \, g$ in $4 \, minutes$, the half-life $T_{1/2} = 4 \, minutes$.We use the radioactive decay formula: $M = M_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}$, where $M$ is the final mass, $M_0$ is the initial mass, and $t$ is the time elapsed.Substituting the values $M = 10 \, g$, $M_0 = 80 \, g$, and $T_{1/2} = 4 \, minutes$:$10 = 80 \left( \frac{1}{2} \right)^{t / 4}$$\frac{10}{80} = \left( \frac{1}{2} \right)^{t / 4}$$\frac{1}{8} = \left( \frac{1}{2} \right)^{t / 4}$Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$, we have:$\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{t / 4}$Equating the exponents: $3 = \frac{t}{4}$$t = 12 \, minutes$.

If $20 \, g$ of a radioactive substance reduces to $10 \, g$ due to radioactive decay in $4 \, minutes$, then in what time will $80 \, g$ of the same substance reduce to $10 \, g$?

Similar Questions

At a certain moment,the amounts of two radioactive substances are in the ratio $2:1$. Their half-lives are $12 \text{ hours}$ and $16 \text{ hours}$ respectively. What will be the ratio of their remaining amounts after $2 \text{ days}$?

$A$ radioactive element has a rate of disintegration of $9000$ disintegrations per minute at a particular instant. After $2$ minutes, it becomes $3000$ disintegrations per minute. The decay constant per minute is:

The half-life of ${}^{198} {Au}$ is $3 \, \text{days}$. If the atomic weight of ${}^{198} {Au}$ is $198 \, \text{g/mol}$, then the activity of $2 \, \text{mg}$ of ${}^{198} {Au}$ is ..... $\times 10^{12} \, \text{disintegration/second}$.

The decay constant $\lambda$ of a radioactive sample is the probability of decay of an atom in unit time. Then,

The activity of a radioactive sample is $6000 \, dps$ after $280$ days. After another $140$ days,the activity decreases to $3000 \, dps$. The initial activity of the radioactive sample (in $dps$) is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module