C32 = 1 2 1
1 2 3 2
N (0)
( )( )
C33 = 1 2 1
1 3 2 3
N (0)
( )( )
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Examples
Example 1: Calculate the activity of K40 in 100 kg man assuming that 0.35% of the body weight is potassium. The abundance of K10 is 0.012%, its half life is 1.31 x 109 years.
Solution:
Total mass of potassium is 100 kg man
= 100 x 0.35 x 10-2 = 0.35 kg
Mass of K40 = 0.350 x 0.012 x 10-2 = 4.20 x 10-5 kg
From Avogadro’s hypothesis, 1 kg atom of a substance consists of 6.023 x 1023 atoms.
Hence, the number of K40 atoms is given by
= 1.061x104 disintegrations/sec
=0.287 micro-curie Ans.
Example 2: If a radioactive material initially contains 3.0 milli grams of uranium (U234), how much it will contain after 150000 years? What will be its activity at the end of this time? (T1/2 = 2.50 x 105 years, = 8.80 x 104 per sec).
Solution:
Let m be the mass of uranium U234 remaining after 150000 years
Number of atoms in mass m = N = m x 6.023 x 1023 / 234 No = 3 x 10-6 x (6.023 x 1023 / 234)
Substituting the numerical values in equation N = Noe-t, we have
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26 26
6 5
mx6.023x10 3x10 x6.023x10 e (0.693 / 2.5x10 )x150000
234 234
or m = 3 x 10-6 e-0.416 0.416 = loge (3 x 10-6/m) or m = 1.98 x 10-6 kg Ans.
Activity of uranium at the end of 150000 years = dN N dt
= 8.8 x 10-14 x 1.98 x 10-6 x 6.023 x 1023/234
= 4.5 x 105 disintegrations/sec
We know that one mole contains Avogadro number of atoms.
Number of atoms in mass (m) = N = Initial number of atoms = N0 =
N = N0 e-λt
=
m = 1.98 × 10-6 kg Ans.
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Example 3: A small quantity of solution containing Na24 radionuclide (half life 15 hour) of activity 1.0 micro-curie is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute.
Determine the total volume of blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. (1 curie = 3.7 × 1010 disintegrations per second).
Solution: We know that λ =
Let the number of radioactive nuclei present after 5 hours be N1 in 1 cm3 sample of blood.
Example 4: A radionuclide with half life T, is produced in a reactor at a constant rate p nuclei per second. During each decay, energy E0 is released. If production of
radionuclide is started at t = 0. Calculate
(a) rate of release of energy as a function of time (b) total energy released upto time t.
Solution: (a) Let at any instant, the number of nuclei in the radionuclide be N.
Rate of decay = λ N
At any instant t, net rate of accumulation of nuclei
= p - λ N
∴ = On integrating both sides, we get
The rate of release of energy at any time t is
A E0 = p E0 [1 – e-(t ln 2/T)
] Ans.
(b) Total number of nuclei produced upto time t = pt Nuclei decayed upto time t = (pt – N)
Example 5: A radioactive source in the form of a metallic sphere of radius 10-2 m emits β –particles at the rate of 5 × 1010 particles per second. The source is electrically insulated.
How long it take for its potential to be raised by 2 volts, assuming that 40% of emitted β –particles escape the source.
Solution: Let t be the time required to raise to potential by 2 V.
Then number of β –particles emitted in t second is 5 × 1010 t.
Now the number of β –particles escaping from sphere is 40%
i.e. 2 × 1010 t. So, the charge developed Q = (2 × 1010 t) (1.6 × 10-19) Coulomb = (3.2 × 10-9 t) Coulomb
But Q = (40 R) V
= (8.85 × 1012) × (10-2) (2) = 17.6 × 10-13
∴ 17.6 × 10-13 = 3.2 × 10-9 t
or t = = 5.53 × 10-5 sec Ans.
Example 6: In an ore containing uranium, the ratio of U238 to Pb206 nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U238. Take the half life of U238 to be 4.5 × 109 years.
Solution: Given that U238 : Pb206 = 3:1
Let us assume that the number of Pb206 nuclei is x, then the number of U238 nuclei will be 3x (= N). Assuming that all the Pb206 present in the ore is due to U238, the initial number of U238 nuclei will be
3x + x = 4x (= N) We know that N = N0 e-λt
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∴ 3x = 4x e-λt or eλt = =
or λ t = ln
∴ t =
λ × ln = 1.868 × 109 years Ans.
Example 7: The element curium 96Cm248 has a mean life of 1013 seconds. Its primary decay modes are spontaneous fission and a decay, the former with a probability of 8%
and the latter with the probability of 92%. Each fission releases 200 MeV of energy. The masses involved in -decay are as follows:
96Cm248 = 248.072220 u
94Pu214 = 244.064100 u And 2He4 = 4.002603 u
Calculate the power output from a sample of 1020 Cd atoms.
Solution: Number of atoms undergoing fission
Nf = × 1020 Number of atoms undergoing -decay
N = × 1020 Energy released in fission process
Ef = Nf × 200 MeV = × 200 = 16× 1020 MeV
Energy released in -decay process
E = Mass defect × 931 × N
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= 0.005517 × 931 × × 1020 = 4.725 × 1020 MeV
Total energy released, E = Ef + E
E = 16 × 1020 + 4.725 × 1020
= 20.725× 1020 MeV
Power output P =
= Watt = 3.3 × 1020 Watt Ans.
Example 8: A nucleus at rest undergoes a decay emitting an - particle of de- Broglie wavelength = 5.76 × 10-15 m. If the mass of the daughter nucleus is 223.610 a.m.u. and that of the - particle is 4.002 a.m.u., determine the total kinetic energy in the final state. Hence, obtain the mass of the parent nucleus in a.m.u. (1 a.m.u. = 931.470 MeV/c2).
Solution: From de-Broglie relation, momentum of -particle is given by
P =
λ ………. (i) COLM gives Pd = P
K.E. is given by K = K + Kd =
+
……… (ii) Where d denotes daughter nucleus
K =
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From equation (i) & (ii), we get
K = λ
Substituting the values, we get
K = 6.25 MeV Ans.
Further, (mp - m - md) c2 = K
∴ (mp – 223.61 – 4.002) c2 = 6.25 mp – 223.61 – 4.002 = 6.25 MeV/c2
mp = 227.62 a.m.u Ans.
Example 9: A thermonuclear device consists of a torus of diameter 3 m with a tube of diameter 1 m, containing deuterium gas at 10-2 mm mercury pressure and at room temperature. A bank of capacitors of 1200 F is discharged through the tube at 40 kV. If only 10% of the electrical energy is transformed to plasma kinetic energy, what is the maximum temperature attained? Assuming that the energy is equally shared between the deuterons and electrons in the plasma.
Solution:
Cross-sectional area of the torus = m2 Circumference = 3 m
Volume of Torus = 32/4 = 7.4 m2
Pressure of the gas = 10-5 x 13.6 x 103 x 9.81 = 1.34 N/m2 From the equation PV = NkT
We have Nk = 1.34 x 7.4/293 = 0.0338 The energy obtained from the discharge
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= CV2 = x 1200 x 10-6 x (4 x 104) = 9.6 x 105 J
Energy transformed to plasma K.E. = 9.6 x 105 J
We know that the average kinetic energy associated with the gas molecule is given as E = NkT
Since, each deuterium molecule produces two ions and electrons, hence 4 NkT = 9.6 x 104
or T = 4.75 x 105 K. Ans.
Example 10: Calculate the energy released by the fission of 2 gm of 92U235 in kWh. Given that the energy released per fission is 200 MeV.
Solution:
The number of atoms in 2 gm of 92U235
=
atoms
Energy released per fission = 200 MeV
= 200 × 1.6 × J = 3.2 × J
Example 11: Assuming that 200 MeV of energy is released per fission of uranium atom find the number of fission per second required to release one kilowatt power.
Solution:
Energy released per fission = 200 Mev
= (200 × ) × 1.6 × J
= 200 × 1.6 × J The rate at which energy is to be released
= 1 kilowatt = 1000 Joules per sec
= Joules per sec.
∴ Number of fission per second =
=3.125 × fission per second Ans.
Example 12: When thermal neutrons are used to induce the reaction
5B10 + 0n1 3Li7 + 2He4
alpha particles are emitted with an energy of 1.83 MeV. Given the masses of boron, neutron and 2He4 as 10.0167 amu ,1.0084 amu, and 4.00386 amu respectively. What is the mass of 3Li7?
Solution:
The given reaction is
5B10 + 0n1 3Li7 + 2He4 + Q
where ‘Q’ is the energy released in the reaction. This energy appears in the form of kinetic energy of the product particles.
Let v1 and v2be the velocities of 3Li7 and 2He4 after the reaction where m1 and m2 are masses.
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According to the law of conservation of momentum =
Net kinetic energy , Q = +
Q =
= =
∴ Q =
or Q = Or Q =
= 1.83 MeV
Q = 1.83 × MeV
= +
= 3.08 × a.m.u.
Substituting the given masses and value of Q in the given reaction 10.0167 + 1.00894 = 3Li7 + 4.00386 + 3.08 ×
3Li7 = (10.0167 + 1.00386) – (4.00386 – 0.00308)
3Li7 = 7.0187 a.m.u. Ans.
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hf
V
Example 13: A stationary He+ ion emitted a photon corresponding to the first line of the Lyman series. This photon liberates electron from a stationary hydrogen atom in the ground state. Find the velocity of the liberated electron.
Solution: Energy of a photon emitted for transition n2 to n1 is given by Δ E = 13.6 eV ……….. (1)
This transition energy is shared between recoiling helium ion and photon.
From conservation of energy, Δ E = hf + K.E. ……….. (2)
Where K.E. is kinetic energy of recoiling atom and hf is energy of photon.
From conservation of momentum,
Photon = - He ……….. (2)
Here negative sign indicates that He+ ion recoils after emission of photons.
Kinetic energy of recoiling of He+ ion
=
=
=
where m is mass of He+ ion.
From equation (2), Δ E = hf + or h2f2 + (2mc2) hf – 2mc2 ΔE = 0
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H Atom
E = 40.8 eV
+ + Electron
On solving above quadratic for hf, we get hf = Δ
Δ We can neglect Δ
in denominator as mc2 > ΔE
H
+Hence, hf ΔE
For first line of Lyman series n2 = 2 and n1 = 1
Δ E = 13.6 eV = 40.8 eV
Ionization energy of Hydrogen atom is 13.6 eV. Since this is less than energy of incident photons, the excess energy is transformed to kinetic energy to liberated electrons,
mv2 = (40.8 – 13.6) eV = 37.2 eV v = 3.1 × 106 m/s Ans.