Question

For $N={{N}_{0}}{{e}^{-\lambda t}}$ and ${{t}_{2}}>{{t}_{1}}$, the number of nuclei disintegrating between ${{t}_{1}}$ and ${{t}_{2}}$ is:
(A) ${{N}_{0}}=[{{e}^{-\lambda /1}}-{{e}^{-\lambda /2}}]$
(B) ${{N}_{0}}=[{{e}^{-\lambda /2}}-{{e}^{-\lambda /1}}]$
(C) ${{N}_{0}}=[{{e}^{\lambda /2}}-{{e}^{\lambda /1}}]$
(D) None of the above

Answer 1

Hint: We know that radioactivity refers to the particles which are emitted from nuclei as a result of nuclear instability. Because the nucleus experiences the intense conflict between the two strongest forces in nature, it should not be surprising that there are many nuclear isotopes which are unstable and emit some kind of radiation. Instability of an atom's nucleus may result from an excess of either neutrons or protons. A radioactive atom will attempt to reach stability by ejecting nucleons (protons or neutrons), as well as other particles, or by releasing energy in other forms.

Complete step by step answer
Initial amount is $\mathrm{N}_{0}$ so amount after time $\mathrm{t}_{1}$ will be $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}$
And that after time $\mathrm{t}_{2}$ is $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$
So decayed in period between $\mathrm{t}_{1}$ and $\mathrm{t}_{2}$ is nothing but
$\mathrm{N}_{1}-\mathrm{N}_{2}=\mathrm{N}_{0}\left(\mathrm{e}^{-\lambda \mathrm{t}_{1}}-\mathrm{e}^{-\lambda \mathrm{t}_{2}}\right)$ as $\mathrm{N}_{1}>\mathrm{N}_{2}$ …... (1)
On solving this integral, we get
$\tau=1 / \lambda$Therefore, we can summarise the observation
as follows:
$\dfrac{{{t}_{1}}}{2}=\dfrac{(\ln 2)}{\lambda }=\tau \ln 2$
The number of nuclei which decay in
the time interval: t to $t+\Delta t$ is:
$\mathrm{R}(\mathrm{t}) \Delta \mathrm{t}=\left(\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t} \Delta \mathrm{t}\right)$……. (2)
Each of them has lived for time $\mathfrak{t}$.
Hence, the total life of all these nuclei is $t \lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda t} \Delta \mathrm{t}$.
Therefore, from equation (1) and (2)
We, can determine that,
${{N}_{0}}=[{{e}^{-\lambda /2}}-{{e}^{-\lambda /1}}]$

Therefore, option B is the correct answer.

Note: We know that there are four major types of radiation: alpha, beta, neutrons, and electromagnetic waves such as gamma rays. They differ in mass, energy and how deeply they penetrate people and objects. Radiation cannot be detected by human senses. A variety of instruments are available for detecting and measuring radiation. The most common type of radiation detector is a Geiger-Mueller (GM) tube, also called a Geiger counter. Utilizing radiation to combat cancer is an important career, earn your radiation therapy degree and help society. Killing Microbes: Gamma rays successfully kill microbes that cause food to decay. So food treated with this radiation has a longer shelf life.