\(\frac { { N }_{ 0 } }{ N } ={ 2 }^{ t/T }\)
(or) 10² = 2^t/1600
\(\frac { 1 }{ { 10 }^{ -2 } } ={ 2 }^{ t/1600 }\)
(or) \(2{ log }_{ 10 }10=\frac { t }{ 1600 } { log }_{ 10 }2\)
(or) \(t=\frac { 2\times 1600 }{ { log }_{ 10 }2 } =\frac { 3200 }{ 0.3010 } =10631.2\) = years

Let N_o, be the number of uranium nuclides present at the initial moment of time and N_u and N_Pb be the number of uranium and lead nuclides present.
Now \(\frac { { N }_{ u } }{ { N }_{ 0 } } { =\left( \frac { 1 }{ 2 } \right) }^{ t/{ T }_{ 1/2 } }={ \left( \frac { 1 }{ 2 } \right) }^{ 1/3 }=\frac { 1 }{ 1.259 } \)
\(\frac { { Nu } }{ { N }_{ u }+{ N }_{ Pb } } =1.259\)
or \(\frac { Nu }{ { N }_{ u }+{ N }_{ Pb } } =1.259\)
or \(\frac { Nu }{ { N }_{ u }+{ N }_{ Pb } } -1=1.259-1\)
or \(\frac { { N }_{ Pb } }{ { N }_{ u } } =0.259\) 

A = Nλ and \(\lambda =\frac { { log }_{ e }2 }{ { T }_{ 1/2 } } \)
or N = \(\frac { A }{ { log }_{ e }2 } { T }_{ 1/2 }\)
Number of moles, n = \(\frac { N }{ { N }_{ A } } \)
\(=\frac { 5\times { 10 }^{ -3 }\times 3.7\times { 10 }^{ 10 }\times 138\times 24\times 3600 }{ 0.693\times 6.023\times { 10 }^{ 23 } } \)
= 5.28 x 10^-9
Mass of Po = nM = 5.28 x 10^-9 x 210 g
= 1.1 x 10^-6 g = 1.1 μg.

The initial number of molecules, N_o = 150
The final number of molecules, N = 150
\(\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n } \ or \ \frac { 150 }{ 600 } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
\({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }={ \left( \frac { 1 }{ 2 } \right) }^{ n } \ or \ n=2=\frac { t }{ { T }_{ 1/2 } } \)
t = 2 x 600 s = 1200 s
Best of disintegration,
\(R=\frac { dN }{ dt } =-\lambda N\)
\(=\frac { 0.693 }{ { T }_{ 1/2 } } \times 150\)
\(=\frac { 0.693 }{ { 600} } \times 150\) = 0.173
disintegration/second at the instant when 150 molecules were remaining.

Let the initial number of nuclei X = N_o
Number of nuclei of X left at any time t = N_x
Number of nuclei of Y at any time t = N_o-N_x = N_y
Now, N_X = N₀e^-λt
N_Y = N₀ - N_X = N_oe^-λt
It is given that \(\frac { { N }_{ X } }{ { N }_{ Y } } =\frac { 1 }{ 15 } \)
\(\frac { { N }_{ o }{ e }^{ -\lambda t } }{ { N }_{ o }({ 1-e }^{ -\lambda t }) } =\frac { 1 }{ 15 } \ or \ { e }^{ -\lambda t }=1\)
e^-λt = 16
\(\lambda t={ log }_{ e }2^{ 4 }=4\times { log }_{ e }2=4\times 0.693\)
\(t=4\times \frac { 0.693 }{ \lambda } \)
\(t=4\times { T }_{ 1/2 }=4\times 50=200\) years

The fusion reaction is as under:
\(_{ 1 }^{ 2 }{ H+ }_{ 1 }^{ 2 }{ H\rightarrow }_{ 2 }^{ 4 }{ He }+Q\) (energy)
Deuteron contains 2 nucleons. Binding energy per nucleon is 1.1 MeV. Binding energy of each deuteron nucleus is 2 x 1.1 i.e. 2 .2 MeV
Total binding energy before reaction
= 2 x 2.2 MeV = 4.4 MeV
Total binding energy after reaction
= 4 x 7 MeV = 28 MeV
Clearly the energy released is (28 - 4.4) MeV
i.e. 23.6. MeV

Using \(N={ N }_{ 0 }{ e }^{ -\lambda t }\), we get
\(3=4{ e }^{ -\lambda t }\) or \(\frac { 4 }{ 3 } ={ e }^{ -\lambda t }\)
or In 4 - In 3 = λt
or \(t=\frac { ln4-ln3 }{ \lambda } =\frac { ln4-ln3 }{ 0.6931 } T\)
\(=\frac { 1.38629-1.09861 }{ 0.6931 } \times 4.5\times { 10 }^{ 9 }yr\)
= 1.868 x 10⁹ yr

The reactor produces 1000 MW power or 10⁹ W power or 10⁹ Js^-1 of power. The reactor is to function for 10 years. Therefore, total energy which the reactor will supply in 10 years is
E = (Power) (time)
= (109 Js^-1) (10 x 365 x 24 x 3600 s)
= 3.1536 x 10¹⁷ J.
But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or 3.1536 x 10¹⁸ J. One uranium atom liberates 200 MeV of energy or 200 x 1.6 x 10-¹³ J or 3.2 x 10^-11 J of energy. So number of uranium atoms needed are
\(\frac { 3.1536\times { 10 }^{ 18 } }{ 3.2\times { 10 }^{ -11 } } =0.9855\times { 10 }^{ 29 }\)
or number of kg-moles of uranium needed are
\(n=\frac { 0.9855\times { 10 }^{ 29 } }{ 6.20\times { 10 }^{ 26 } } =163.7\)
Hence total mass of uranium required is
m = (n) M = (163.7) (235) kg
or m = 38470 kg.
or m = 3.847 x 10⁴ g

Strength of radioactive source
= 8.0 mCi = 8.0 x 10^-3 Ci
= 8.0 x 10^-3 x 3.7 x 10¹⁰ disintegrations s^-1
Since the strength of the source decreases with time,
\(\therefore \frac { dN }{ dt } =-29.6\times { 10 }^{ 7 }\)
But \(\frac { dN }{ dt } =-\lambda N\)
\(\therefore =-\lambda N=-29.6\times { 10 }^{ 7 }\)
or \(\lambda N=-29.6\times { 10 }^{ 7 }N=\frac { 29.6\times { 10 }^{ 7 } }{ \lambda } \)
or \(N=\frac { 29.6\times { 10 }^{ 7 }\times T }{ 0.693 } \)

The atomic weight of lithium
= \(\frac { 7.5\times 6.01512+92.5\times 7.01600 }{ 100 } \)
\(=\frac { 45.1134+648.98 }{ 100 } =6.94093\)

\(N=\frac { 6.023\times { 10 }^{ 23 } }{ 90 } \times 15\times { 10 }^{ -3 }\)
\(\frac { dN }{ dt } =\lambda N=\frac { 0.693 }{ { T }_{ 1/2 } } N\)
= \(\frac { 0.693 }{ 28\times 365\times 24\times 60\times 60 } \times \frac { 6.023{ 10 }^{ 23 } }{ 90 } \times 15\times { 10 }^{ -3 }Bq\)
= 7.878 x 10¹⁰ Bq.

Chain reaction :

(i) When one \({ }_{92}^{235} \mathrm{U}\) Unucleus undergoes fission,the energy released might be small. But from each fission reaction, three neutrons are released.
(ii) These three neutrons can cause further fission in three other \({ }_{92}^{235} \mathrm{U}\) nuclei which in turn produce nine neutrons. These nine neutrons initiate fission in another 27 \({ }_{92}^{235} \mathrm{U}\) nuclei and so on.
(iii) This process is called a chain reaction and the number of neutrons goes on increasíng almost in geonetric progression. There are two kinds of chain reactions :
(a) uncontrolled chain rcaction               
(b) controlled chain reaction.
(a) uncontrolled chain rcaction :
The number of neutrons multiplies indefinitely and the entire amount of energy released in a fraction of second.
Example : Atom bomb.
(b) controlled chain reaction :
The average number of neutrons released in each stage is kept as one such that it is possible to store the released energy.
Example : Nuclear reactor.

(i) The smoke detector uses around 0.2 mg of a man-made weak radioactive isotope called americium \((_{ 95 }^{ 241 }{ Am })\)

(ii) This radioactive source is placed between two oppositely charged metal plates and α radiations from \(_{ 95 }^{ 241 }{ Am }\) continuously ionize the nitrogen, oxygen molecules in the air space between the plates
(iii) As a result, there will be a continuous flow of small steady currents in the circuit.
(iv) If smoke enters, the radiation is being absorbed by the smoke particles rather than air molecules.
(v) As a result, the ionization and along with it the current is reduced. This drop-in current is detected by the circuit and the alarm starts.
(vi) The radiation dosage emitted by americium is very much less than the safe level, so it can be considered harmless.

Rutherford atom model helps in the calculation of the diameter of the nucleus and also the size of the atom but has the following limitations
(a) This model fails to explain the distribution of electrons around the nucleus and also the stability of the atom.
According to classical electrodynamics,.any accelerated charge emits electromagnetic radiations. Due to emission of radiations, it loses its energy. Hence, it can no longer .sustain the circular motion. The radius of the orbit, therefore, becomes smaller and smaller (undergoes spiral motion) and finally the electron should fall into the nucleus and the atoms should disintegrate. But this does not happen.

(b) According to this model, emission of radiation must be continuous and must give continuous emission spectrum but experimentally we observe only line (discrete) emission spectrum for atoms.

(i) In 1911, Geiger and Marsden did a remarkable experiment based on the advice of their teacher Rutherford, which is known as the scattering of alpha particles by gold foil.

(ii) The experimental arrangement. A source of alpha particles (radioactive material, for example, polonium) is kept inside a thick lead box, with a fine hole.
(iii) The alpha particles coming through the fine hole of the lead box pass through another fine hole made on the lead screen. These particles are now allowed to fall on a thin gold foil and it is observed that the alpha particles passing through gold foil are scattered through different angles.
(iv) A movable screen (from 0° to 180°) which is made up of zinc sulphide (ZnS) is kept on the other side of the gold foil to collect the alpha particles. Whenever alpha particles strike the screen, a flash of light is observed which can be seen through a microscope.
(v) Rutherford proposed an atom model based on the results of alpha scattering. experiment.
(vi) In this experiment, alpha particles (positively charged particles) are allowed to fall on the atoms of a metallic gold foil. The results of this experiment. Rutherford expected the nuclear model, but the experiment showed the model.

(a) Most of the alpha particles were un-deflected through the gold and went straight.
(b) Some of the alpha particles are deflected through a small angle.
(c) A few alpha particles (one in a thousand) are deflected through an angle more than 90°.
(d) Very few alpha particles returned back (backscattered) that is, deflected back by 180°.