Question:
The decay constant of a radio isotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1 $ and $t_2$ respectively, the number of nuclei which have decayed during the time $(t_1-t_2)$
The decay constant of a radio isotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1 $ and $t_2$ respectively, the number of nuclei which have decayed during the time $(t_1-t_2)$
Updated On: Jul 18, 2024
- $A_1t_1-A_2t_2$
- $A_1-A_2$
- $(A_1-A_2)/ \lambda$
- $\lambda (A_1-A_2)$
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The Correct Option is C
Solution and Explanation
$A_1=\lambda N_1 \, at \, time \, t_1$
$A_2=\lambda N_2$ at time $t_2$
Therefore, number of nuclei decayed during time interval $(t_1-t_2)$ is
$N_1-N_2 =\frac{[A_1-A_2]}{\lambda}$
$A_2=\lambda N_2$ at time $t_2$
Therefore, number of nuclei decayed during time interval $(t_1-t_2)$ is
$N_1-N_2 =\frac{[A_1-A_2]}{\lambda}$
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Concepts Used:
Decay Rate
The disintegration of unstable heavy atomic nuclei into lighter, more stable, atomic nuclei, accompanied in the process by the emission of ionizing radiation (alpha particles, beta particles or gamma rays). This is a random process at the atomic level but, given a large number of similar atoms, the decay rate on average is predictable, and is usually measured by the half-life of the substance.
The equation for finding out the decay rate is given below:
