2 Marks Questions

Question 12 Marks

Define the term 'Half-life' of a radioactive substance. Two different radioactive substances have half-lives $T _1$ and $T _2$ and number of undecayed atoms at an instant, $N _1$ and $N _2$, respectively. Find the ratio of their activities at that instant.

Answer

Let $R_1$ and $R_2$ be their activities then
$R _1=\lambda_1 N_1$
$R _2=\lambda_2 N_2$
$\frac{ R _1}{ R _2}=\frac{\lambda_1 N_1}{\lambda_2 N_2}=\frac{\frac{ N _1}{T_1}}{\frac{N_2}{T_2}}=\frac{ N _1 T_2}{N_2 T_1}$

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Question 22 Marks

Write change in position of atom of an element in periodic table due to emission of $\alpha$ and $\beta$ particles from its nucleus.

Answer

Atomic number of radioactive element reduced by 2 due to emission of 1-$\alpha$-particle, hence the position of the new atom lags behind 2 period in periodic table.
Atomic number of radioactive element increases by 1 due to emission of 1-$\beta$-particle, hence the position of the new atom leads ahead by 1 period in periodic table.

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Question 32 Marks

What is the importance of binding energy per nucleon ?

Answer

Binding energy per nucleon
$ = \frac{\\text{Total binding energy}}{\text{Atomic mass number}} $.
It is the measure of stability of a nucleus. Greater is the binding energy per nucleon of a nucleous, much stable is the nucleus i.e. much external energy is required to decompose the nucleus into its nucleons.

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Question 42 Marks

Half-life time of a radioactive element is 10 years. In how much time it will be left (1/16)th of its original mass ?

Answer

Given: $ T = 10 \text{ years} $; $ m = m_{0}/16 $, where $m$ = original mass of radioactive element.
$\therefore \quad N = N _0\left(\frac{1}{2}\right)^n \Rightarrow m=m_0\left(\frac{1}{2}\right)^n$
$\Rightarrow \quad \frac{m_0}{16}=m_0\left(\frac{1}{2}\right)^n$
$\Rightarrow \quad\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^n \Rightarrow n=4$
∴ Time of radioactive decay $t=n \times T=4 \times 10$ years
= 40 years

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