Question

In an experiment, the initial number of radioactive nuclei is $3000$. It is found that $1000\pm 40$ nuclei decayed in the first $1.0\sec$. For $\left|\mathrm{x}\right|<<1$, $\mathrm{In}(1+\mathrm{x})=\mathrm{x}$ up to the first power in $\mathrm{x}$. The error $\mathrm{\Delta \lambda }$, in the determination of the decay constant $\mathrm{\lambda }$, in ${\mathrm{s}}^{-1}$ is

• A. $0.04$
• B. $0.03$
• C. $0.02$
• D. $0.01$

Answer 1

The correct option is C

$0.02$

Step 1: Given data

In the question statement, it is given that:

For$\left|\mathrm{x}\right|<<1$, $\mathrm{In}(1+\mathrm{x})=\mathrm{x}$ up to the first power in $\mathrm{x}$

The initial number of radioactive nuclei $=3000$

Step 2: Formula used to find decay constant:

Number of nuclei decay in time $t$ $={\mathrm{N}}_0–{\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}$

$\mathrm{N}={\mathrm{N}}_0–{\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}$ ,where $\mathrm{N}$ is the number of decayed nuclei, ${\mathrm{N}}_0$ is the number of initial nuclei, $\mathrm{\lambda }$ is decay constant, e is a mathematical constant, and $\mathrm{t}$ is time.

$\mathrm{dN}=0–{\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}(-\mathrm{d\lambda })$, where $\mathrm{dN}$ is the number of nuclei decayed at an instant time.

$\mathrm{d\lambda }=\frac{\mathrm{dN}}{{\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}}$

Step 3: Find decay constant by putting value,

$\mathrm{d\lambda }=\left(\frac{40}{3000}\right)(1–\mathrm{\lambda t})$

$1000=3000–3000.{\mathrm{e}}^{-\mathrm{\lambda }1}$

$\frac{2}{3}={\mathrm{e}}^{-\mathrm{\lambda }1}$

$\Rightarrow \mathrm{d\lambda }=\frac{(40\mathrm{x}3)}{(3000\mathrm{x}2)}=0.02$

Thus, the decay constant is $0.02$.

Hence, option C is correct.