Radioactivity of a sample (Z=22) decreases by | vedclass

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

Question

(C) The radioactive decay follows first-order kinetics. The decay constant $\lambda$ is given by: $\lambda = \frac{2.303}{t} \log \frac{N_0}{N_t}$.Giv

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The radioactive decay follows first-order kinetics. The decay constant $\lambda$ is given by: $\lambda = \frac{2.303}{t} \log \frac{N_0}{N_t}$.Given that radioactivity decreases by $90 \%$, the remaining amount $N_t = N_0 - 0.90 N_0 = 0.10 N_0$.Substituting the values: $\lambda = \frac{2.303}{10} \log \frac{N_0}{0.10 N_0} = \frac{2.303}{10} \log(10) = \frac{2.303}{10} 	imes 1 = 0.2303 \ year^{-1}$.The half-life $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{\lambda}$.$t_{1/2} = \frac{0.693}{0.2303} \approx 3.01 \ years$.Therefore, the half-life is approximately $3 \ years$.

Radioactivity of a sample $(Z=22)$ decreases by $90 \%$ after $10 \ years$. What will be the half-life of the sample (in $years$)?

Similar Questions

$A$ certain nuclide has a half-life period of $30 \ minutes$. If a sample containing $600$ atoms is allowed to decay for $90 \ minutes$,how many atoms will remain?

After $2 \, \text{hours}$,the amount of a certain radioactive substance reduces to $1/16^{th}$ of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is $...... \, \text{min}$.

The half-life period of ${}_{53}I^{125}$ is $60$ days. The radioactivity after $180$ days will be (in $\%$)

$A$ radioactive isotope having $t_{1/2} = 3 \ days$ was measured after $12 \ days$. If $3 \ g$ of the isotope remains in the container,what was the initial weight of the isotope (in $g$)?

What type of reaction order is followed by radioactive processes?

Vedclass Test Series

Exam Paper Generator

Online Exam Module