Question

In an experiment the initial number of radioactive nuclei is $3000$. It is found that $1000\pm 40$ nuclei decayed in the $1.0s$, for $\left| x \right|\ll 1,\text{ }\ln \left( 1+x \right)=x$ up to first power in $x$. The error $\Delta \lambda $, in the determination of the decay constant, in ${{s}^{-1}}$, is:
$\begin{align}
  & \text{A}\text{. }0.04 \\
 & \text{B}\text{. }0.03 \\
 & \text{C}\text{. }0.02 \\
 & \text{D}\text{. }0.01 \\
\end{align}$

Answer 1

Hint: The process by which an unstable atomic nucleus loses energy by radiation is called radioactive decay. The radioactive decays per unit time are directly proportional to the number of nuclei of radioactive compounds in the sample. We will use the expression for radioactive decay relating the number of decayed nuclei to the initial number of nuclei and the decay constant.

Formula used:
Number of decayed nuclei,
$N={{N}_{o}}-{{N}_{o}}{{e}^{-\lambda t}}$

Complete step by step answer:
Radioactive decay is described as the process by which an unstable atomic nucleus loses energy by radiation. A sample material containing radioactive nuclei is considered as radioactive. The decay of radioactive elements occurs at a fixed constant rate.

According to the radioactive decay law, the radioactive decays per unit time are directly proportional to the number of nuclei of radioactive compounds in the sample. Let the number of nuclei in a sample is $N$ and the number of radioactive decays per unit time $dt$ is $dN$, then,

$\dfrac{dN}{dt}=-\lambda N$
Where,
$\lambda $ is the decay constant

Decay constant is defined as the proportionality between the size of population of radioactive atoms and the rate at which this population decreases as a result of radioactive decay. Population of radioactive atoms refers to the number of radioactive nuclei.
Number of nuclei decayed in particular time being is given as,

$N={{N}_{o}}-{{N}_{o}}{{e}^{-\lambda t}}$
Where,
$N$ is the number of decayed nuclei
${{N}_{o}}$ is the number of initial nuclei
$\lambda $ is the decay constant
$t$ is the time

Now,
$\begin{align}
  & dN=0-{{N}_{o}}{{e}^{-\lambda t}}\left( -d\lambda \right) \\
 & dN={{N}_{o}}{{e}^{-\lambda t}}d\lambda \\
\end{align}$
Where,
$dN$ is the number of nuclei decayed in an instant of time
$d\lambda =\dfrac{dN}{{{N}_{o}}{{e}^{-\lambda t}}}$

We have,
$\begin{align}
  & {{N}_{o}}=3000 \\
 & dN=40 \\
\end{align}$

Therefore,

$d\lambda =\dfrac{40}{3000}\left[ 1-\lambda t \right]$
$N={{N}_{o}}-{{N}_{o}}{{e}^{-\lambda t}}$
Put,
$\begin{align}
  & {{N}_{o}}=3000 \\
 & N=1000 \\
 & t=1 \\
\end{align}$

We get,

$\begin{align}
  & 1000=3000-3000{{e}^{-\lambda \times 1}} \\
 & \dfrac{2}{3}={{e}^{-\lambda \times 1}} \\
\end{align}$
As,
$d\lambda =\dfrac{dN}{{{N}_{o}}{{e}^{-\lambda t}}}$
Therefore,
$\begin{align}
 & d\lambda =\dfrac{40\times 3}{3000\times 2}=\dfrac{2}{100} \\
 & d\lambda =0.02 \\
\end{align}$
The error $\Delta \lambda $, in the determination of the decay constant is $0.02$
Hence, the correct answer is option C.

Note:
We can also use the below approach to solve the given question:
We have,

$N={{N}_{o}}-{{N}_{o}}{{e}^{-\lambda t}}$
$dN={{N}_{o}}{{e}^{-\lambda t}}d\lambda $
Where,
$N$ is the number of decayed nuclei
${{N}_{o}}$ is the number of initial nuclei
$\lambda $ is the decay constant
$t$ is the time
$dN$ is the number of nuclei decayed in an instant of time

Converting the equation into fractional error equation of number of radioactive nuclei, we get,
$\begin{align}
  & \dfrac{\Delta N}{N}=\Delta \lambda t \\
 & \Delta \lambda =\dfrac{\Delta N}{Nt} \\
\end{align}$

Given that,
$\begin{align}
  & N=2000 \\
 & \Delta N=40 \\
 & t=1s \\
\end{align}$
Therefore,
$\begin{align}
  & \Delta \lambda =\dfrac{40}{2000\times 1}=\dfrac{2}{100} \\
 & \Delta \lambda =0.02 \\
\end{align}$
Hence, the error $\Delta \lambda $, in the determination of the decay constant is $0.02$