The radioactivity of a radioactive element be | vedclass

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

Question

(C) The decay constant $\lambda$ is given by the formula: $\lambda = \frac{2.303}{t} \log \left( \frac{N_0}{N} \right)$.Given $t = 2.303 \ s$ and $\fr

Vedclass · Accepted Answer

$0.693$

Vedclass · Answer

(C) The decay constant $\lambda$ is given by the formula: $\lambda = \frac{2.303}{t} \log \left( \frac{N_0}{N} ight)$.Given $t = 2.303 \ s$ and $\frac{N}{N_0} = \frac{1}{10}$,so $\frac{N_0}{N} = 10$.Substituting the values: $\lambda = \frac{2.303}{2.303} \log(10) = 1 	imes 1 = 1 \ s^{-1}$.The half-life period $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{\lambda}$.Therefore,$t_{1/2} = \frac{0.693}{1} = 0.693 \ s$.

The radioactivity of a radioactive element becomes $\frac{1}{10}$ of the original radioactivity after $2.303 \ s$. The half-life period is: (in $s$)

Similar Questions

How many alpha particles are emitted per second by $1 \ \mu g$ of radium?

$C-14$ is used in carbon dating of dead objects because

Decay of $_{92}U^{235}$ is $...$ order reaction.

Half-life of a radioactive substance which disintegrates by $75\%$ in $60 \ min$ will be ......... $min$.

The half-life of a radioactive substance is $120$ days. After $480$ days,the amount of $4 \ g$ of the substance remaining will be: (in $g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module