Radioactivity MCQ Questions & Answers in Modern Physics | Physics

Let $${N_0}$$ be the number of atoms of $$X$$ at time $$t=0.$$
Then at $$t = 4\,hrs$$   (two half lives)
$$\eqalign{ & {N_x} = \frac{{{N_0}}}{4}\,\,{\text{and}}\,\,{N_y} = \frac{{3{N_0}}}{4} \cr & \therefore \frac{{{N_x}}}{{{N_y}}} = \frac{1}{3} \cr} $$
and at $$t = 6\,hrs$$   (three half lives)
$${N_x} = \frac{{{N_0}}}{8}\,\,{\text{and}}\,\,{N_y} = \frac{{7{N_0}}}{8}\,\,{\text{or}}\,\,\frac{{{N_x}}}{{{N_y}}} = \frac{1}{7}$$
The given ratio $$\frac{1}{4}$$ lies between $$\frac{1}{3}$$ and $$\frac{1}{7}.$$
Therefore, $$t$$ lies between $$4\,hrs$$  and $$6\,hrs.$$

$$\eqalign{ & {N_1} = {N_0}{e^{ - 10\lambda t}}{\text{ and }}{N_2} = {N_0}{e^{ - \lambda t}} \cr & \therefore \frac{{{N_1}}}{{{N_2}}} = \frac{{{e^{ - 10\lambda t}}}}{{{e^{ - \lambda t}}}} = \frac{1}{{{e^{9\lambda t}}}} \cr & {\text{Given}}\,\frac{{{N_1}}}{{{N_2}}} = \frac{1}{e};\,\,\,\,\therefore \frac{1}{{{e^{9\lambda t}}}} = \frac{1}{e} \cr & {\text{or,}}\,9\lambda t = 1\,\,\,\,{\text{or}}\,t = \left( {\frac{1}{{9\lambda }}} \right) \cr} $$

$$_1{H^2}{ + _1}{H^2}{ \to _2}H{e^4}$$
Total binding energy of two deuterium nuclei $$ = 1.1 \times 4 = 4.4\,MeV$$
Binding energy of a $$\left( {_2H{e^4}} \right)$$  nuclei $$ = 4 \times 7 = 28\,MeV$$
Energy released in this process $$ = 28 - 4.4 = 23.6\,MeV$$

$$_2H{e^4}$$  contains 2 neutrons and 2 protons
So, mass of 2 protons $$ = 2 \times 1.0073 = 2.0146\,u$$
So, mass of 2 neutrons $$ = 2 \times 1.0087 = 2.0174\,u$$
Total mass of 2 protons and 2 neutrons $$ = \left( {2.0146 + 2.0174} \right)u = 4.032\,u$$
Mass of helium nucleus $$ = 4.0015\,u$$
Thus, mass defect is lacking of mass in forming the helium nucleus from 2 protons and 2 neutrons.
$$\eqalign{ & \therefore \Delta m = {\text{mass}}\,{\text{defect}} = \left( {4.032 - 4.0015} \right)u \cr & = 0.0305\,u \cr} $$
As we know that, $$1\,u = 931\,MeV$$
Hence, binding energy $$\Delta E = \left( {\Delta m} \right) \times 931$$
$$\eqalign{ & = 0.0305 \times 931 \cr & = 28.4\,MeV \cr} $$

The activity of a radioactive substance is defined as the rate at which the nuclei of its atoms in the sample disintegrate. In two half-lives, the activity becomes one-fourth. Two months is 2 half-life period. The activity, two months earlier was
$$2 \times {2^2} = {\text{8}}\,{\text{microcurie}}{\text{.}}$$
NOTE
The activity of a radioactive sample is called one curie, if it undergoes $$3.7 \times {10^{10}}$$   disintegrations per second.

A nuclear reaction represents the transformation of one stable nucleus into another nucleus by bombarding the former with suitable high energy particles.
$$_6{C^{12}}{ + _0}{n^1}{ \to _6}{C^{13}}{ \to _7}{N^{13}}{ + _{ - 1}}{\beta ^0} + Q\left( {{\text{Energy}}} \right)$$
Resulting nucleus is of nitrogen having mass no. 13 and atomic no. 7

We have $${K_\alpha } = \frac{{{m_y}}}{{{m_y} + {m_\alpha }}}.Q$$
$$\eqalign{ & \Rightarrow {K_\alpha } = \frac{{A - 4}}{A}.Q \cr & \Rightarrow 48 = \frac{{A - 4}}{4}.50 \cr & \Rightarrow A = 100 \cr} $$

Let $$E$$ be the energy of proton, then
$$\eqalign{ & E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right] \cr & \Rightarrow E = 56.48 - 39.2 = 17.28\,MeV \cr} $$

As a result of emission of $$1\alpha $$ -particle, the mass number decreases by $$4$$ units and atomic number decreases by $$2$$ units. And by the emission of $$1$$ positron the atomic number decreases by $$1$$ unit but mass number remains constant.
∴ Mass number of final nucleus $$= A - 12$$
Atomic number of final nucleus $$= Z - 8$$
∴ Number of neutrons $$ = \left( {A - 12} \right) - \left( {Z - 8} \right)$$
$$ = A - Z - 4$$
Number of protons $$= Z - 8$$
∴ Required ratio $$ = \frac{{A - Z - 4}}{{Z - 8}}$$

According to question,
Half life of $$X,{T_{\frac{1}{2}}} = {\tau _{av}},$$   average life of $$Y$$
$$\eqalign{ & \Rightarrow \frac{{0.693}}{{{\lambda _X}}} = \frac{1}{{{\lambda _Y}}} \Rightarrow {\lambda _X} = \left( {0.693} \right).{\lambda _Y} \cr & \therefore {\lambda _{\text{X}}} < {\lambda _{\text{Y}}}. \cr} $$
Now, the rate of decay is given by
$$ - \frac{{dN}}{{dt}} = \lambda N$$
∴ $$Y$$ will decay faster than $$X.$$ [$$\because N$$  is some]