235 g of \(_{ 92 }^{ 235 }{ U }\) has 6.02 x 10²³ atoms. In one gram of \(_{ 92 }^{ 235 }{ U }\), the number of atoms is equal to \(\frac { 6.02\times { 10 }^{ 23 } }{ 235 } =2.56\times { 10 }^{ 21 }\)
So the number of atoms in 1 kg of \(_{ 92 }^{ 235 }{ U }\) = 2.56 x 10²¹ x 1000 = 2.56 x 10²⁴
Each \(_{ 92 }^{ 235 }{ U }\) nucleus releases 200 MeV of energy during the fission. The total energy released by 1kg of \(_{ 92 }^{ 235 }{ U }\) is
Q = 2.56 x 10²⁴ x 200MeV = 5.12 x 10²⁶ MeV
In terms of joules,
Q = 5.12 x  10²⁶ x 1.6 x 10^-13 J = 8.192 x 10¹³ J
In terms of Kilowatt hour,
Q = \(\frac { 8.192\times { 10 }^{ 13 } }{ 3.6\times { 10 }^{ 6 } } =2.27\times { 10 }^{ 7 }\) kWh

Energy produced per second in reactor = 1 W = 1 J/s
Energy produced per fission = 200 MeV = 200 x 10⁶ x 1.6 x 10^-19 J
= 3.2 x 10¹¹ J
Number of fissions per second required \(=\frac{\text { Energy produced per second in reactor }}{\text { Energy produced per fission }} \)
\(=\frac{1}{3.2 \times 10^{11}}=\frac{10 \times 10^{10}}{3.2}=3.125 \times 10^{10} \)
Number of fissions per second = 3.125 x 10¹⁰ 

To calculate the age, we need to know the initial activity (R₀) of the caracol (when the sample was alive).
The activity R of the sample
R = R₀ e^-λt   ...(1)
To find the time t, rewriting the above equation (1),
\({ e }^{ \lambda t }=\frac { { R }_{ 0 } }{ R } \)
By taking the logarithm on both sides, we get \(t=\frac { 1 }{ \lambda } In\left( \frac { { R }_{ 0 } }{ R } \right) \) ..(2)
Here R = 38 decays/s = 38 Bq.
To find decay constant, we use the equation
\(\lambda =\frac { 0.6931 }{ { T }_{ 1/2 } } =\frac { 0.6931 }{ 5730yr\times 3.156\times { 10 }^{ 7 }s/yr } \)
[∴ 1yr = 365.25 x 24 x 60 x 60 s = 3.156 x 10⁷ s]
λ = 3.83 x 10⁻¹² s⁻¹
To find the initial activity R₀, we use the equation R₀ = λN₀. Here N₀ is the number of carbon-14 atoms present in the sample when it was alive. The mass of the charcoal is 200 g. In 12 g of carbon, there are 6.02 x 10²³ carbon atoms. So 200 g contains,
\(\frac { 6.02\times { 10 }^{ 23 }atoms/mol }{ 12g/mol } \times 200\approx 1\times { 10 }^{ 25 }\) atoms
When the tree(sample) was alive, the ratio of \(_{ 6 }^{ 14 }{ C }{ e }\) to \(_{ 6 }^{ 12 }{ C }{ e }\) is 1.3 x 10^-12. So the total number of carbon-14 atoms is given by
N₀ = 1 x 10²⁵ x 1.3 x 10^-12 atoms
The initial activity
R₀ = 3.83 x 10^-12 x 1.3 x 10¹³ ≈ 50 decays / s
= 50 Bq
By substituting the value of R₀ and λ in the equation (2), we get
\(t=\frac { 1 }{ 3.83\times { 10 }^{ -12 } } \times In\left[ \frac { 50 }{ 37 } \right] \)
\(t=\frac { 0.301 }{ 3.83 } \times { 10 }^{ 12 }\approx 7.86\times { 10 }^{ 10 }\)sec
In years
\(t=\frac { 7.86\times { 10 }^{ 10 }s }{ 3.156\times { 10 }^{ 7 }s/yr } \approx 2500\) years
In fact, the excavated materials were to USA sent for carbon dating by the Archeological Department of Tamilnadu and the report confirmed that the age of Keezhadi artifacts lies between 2200 years to 2500 years (Sangam era- 400 BC to 200 BC). The Keezhadi excavations experimentally proved that urban civilization existed in Tamil Nadu even 2000 years ago!

λN = 3.9
λN_o = 1.6
\(\frac { N }{ { N }_{ o } } =\frac { 3.9 }{ 15.6 } =\frac { 1 }{ 4 } ={ e }^{ -\lambda t }\)
\({ e }^{ -\lambda t }=\frac { N }{ { N }_{ o } } =4\)
λt = log_e 4
= 2.303 x log₁₀4

A number of fissions occurring/s to give an energy output of 100 MW (= 10⁸ Js^-1).
= \(\frac { { 10 }^{ 8 }{ Js }^{ -1 } }{ 185\times { 10 }^{ 6 }\times 1.6\times { 10 }^{ -19 }J } \)
= 3.378 x10¹⁸ fission s^-1
Number of nuclei contained in 1 kg of \(_{ }^{ 235 }{ U }\)
\(\frac { 1kg }{ 235kg/k \ mol } \times 6.023\times { 10 }^{ 26 }\frac { nuclei }{ k \ mol } \)
= 2.563 x 10²⁴nuclei
Time taken to exhaust 1 kg.of \(_{ }^{ 235 }{ U }\)
= \(\frac { 2.563\times { 10 }^{ 24 } }{ 3.378\times { 10 }^{ 18 } } \) second
= 8.78 days.

(i) Initial decay rate = λN₀
5.6 x 10¹⁰ = 2.1 X 10^-6N_o
or N₀ = \(\frac { 5.6\times { 10 }^{ 10 } }{ 2.2\times { 10 }^{ -16 } } =2.67\times { 10 }^{ 16 }\)
(ii) Activity at any time t is proportional to the number of atoms present at time t
\(N=\frac { { N }_{ 0 } }{ 4 } \)
Now, \(\frac { { N }_{ 0 } }{ 4 } ={ N }_{ 0 }\) or \({ e }^{ -\lambda t }=4\)
Taking logs, λt = log 4 = 2 log_e 2 = 2 x 0.693
\(t=\frac { 2\times 0.693 }{ \lambda } =\frac { 2\times 0.693 }{ 2.1\times { 10 }^{ -6 } } s\)
= 0.66 x 10⁶ s = 6.6 x 10⁵ S

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 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

(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.

(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°.