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Q.1. A nucleus at rest undergoes a decay emitting an α  particle of deBroglie wavelength λ = 5.76 x 10^{15} m. If the mass of the daughter nucleus is 223.610 amu and that of the α  particle is 4.002 amu , determine the total kinetic energy in the final state. Hence, obtain the mass of the parent nucleus in amu.
Let _{Z}X^{4} represent the parent nucleus. The parent nucleus decays emitting α  particles
∴ _{Z}X^{4} → _{Z2} Y^{A4} + _{2}He^{4}Mass of daughter nucleus Y = 223.610 amu , Mass of α  particle = 4.002 amu
DeBroglie wavelength of α  particle, λ = 5.76 x 10^{15} m
So, momentum of α  particle would beFrom law of conservation of linear momentum, this should also be equal to the linear momentum the daughter nucleus thus p_{α} = p_{Y}
Let K_{α} and K_{Y} be the kinetic energies of α  particle and daughter nucleus. Then total kinetic energy in the final state is∵ 1 amu 1.67x10^{27} kg
Mass equivalent to energy
or Δm = 0.0067amu ∴ m_{X} = m_{Y} + m_{α} + Δm
or m_{X} = (223.610 + 4.002 + 0.0067)amu or m_{X} = 227.67∴ Mass of parent nucleus = 227.62 amu
Q.2. A nucleus with mass number 220 initially at rest emits an α  particle. If the Q value of the reaction is 5.5 MeV , calculate the kinetic energy of the α  particle
Linear momentum is conserved.
∴ p_{1} = p_{2}
but where K = kinetic energy
where K_{2} is for α  particle and K_{1} is for nucleus.
or 216K_{1} = 4K_{2} or K_{2} = 54K1 ....(i)
Given: K_{1} + K_{2} = 5.5MeV .....(ii)From (i) and (ii),
K_{1} + 54K_{1} = 5.5 (MeV)
or 55K_{1} = 5.5 (MeV)
∴ K_{2}54 x K_{1} or or K_{2} = 5.4 MeV ....(iv)
∴ Kinetic energy of α  particle = 5.4 MeV
Q.3. In a nuclear reaction ^{235}U undergoes fission liberating 200 MeV of energy. The reactor has 10% efficiency and produces 1000MW power. If the reactor is to function for 10 year, find the total mass of uranium required.
∴ Equivalent energy = Power input x time = 10^{10 }x 365 x 24 x 60 x 60 J
= 31.1536 x 10^{18} J ...(i)Now 1 fission releases 200 MeV energy
∴ 1 fission releases energy = 200 x 1.6 x 10^{13} J (ii)From (i) and (ii),
or Number of fissions = 9.85 x 10^{28}
∴ Number of U^{235} atoms = 9.85 x 10^{28}
Avogadro number 6.02 x 10^{26} per kg 26∴ 6.02 x 10^{6} atoms are contained in 235kg of U^{235}
∴ 9.85 x 10^{28} atoms are contained in mass
∴ Mass of U^{235} = 38451kg
Q.4. If a star can convert all the He nuclei completely into oxygen nuclei, then find the energy released per oxygen nuclei. [Mass of He nucleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu]
4[_{2}He^{4}] →8 O^{16}
Binding energy Δm x 931.5 MeV = (4 x 4.0026 15.9994) x 931.5 MeV = 10.24MeV
∴ Energy released per oxygen nuclei = 10.24 MeV
Q.5. The element Curium has a mean life of 10^{13} second, its primary decay modes are spontaneous fission and αdecay, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV of energy. The masses involved in αdecay are as follows:
Calculate the power output from a sample of 10^{20}Cm atoms.
The primary decay modes of _{96}Cm^{248} are
(i) Spontaneous fission (probability 8%)
(ii)  α decay (probability 92%).The nuclear reaction is given as:
_{96}Cm^{248} → _{94}Pu^{244} + _{2}He^{4}∴ Mass defect in the reaction = Δm
∴ Δm = mass of Cm  mass of Pu  mass of He
or Δm = 248.072220  244.064100  4.002603 or Δm = 0.005517u∴ Energy released = (0.00515 x 931) MeV
E_{α} = 5.136 MeV
Given: = E_{f} Each fission released 200 meV of energyMean life of 10^{13} sec
again dN/dt = λN where N is given to be 10^{20}
∴ Rate of decay = (10^{13})(10^{20}) or Rate of decay 10^{7} dps
of these dps 10^{7} dps, 8% are in fission and 92% are in α  decay process.
Energy released per second due to fission
Energy released per sec due to α  decay
Total energy released per second = (16 + 4.725)10^{7} MeV = 20.725 x 10^{7} MeV
∴ Power output = Energy per second
= (20.725 x 10^{7}) x (1.6 x 10^{13})J / s = 3.316 x 10^{5} W = 3.32 x 10^{5} W
Q.6. A star initially has 10^{40 }deuterons. It produces energy via the processes
_{1}H^{2} + _{1}H^{2} → _{1}H^{3} + p, and _{1}H^{2} + _{1}H^{2} →_{2} He^{4} + n.
If the average power radiated by the star is 10^{16}W , then the time required for the deuteron supply of the star to be exhausted.
The masses of the nuclei are as follow:
M(H^{2}) = 2.014 amu ; M(p) = 1.007 amu; M (n) = 1.008 amu; M (He^{4}) = 4.001amu.
_{1}H^{2} + _{1}H^{2} → _{1}H^{3} + p, and _{1}H^{2} + _{1}H^{2} →_{2} He^{4} + n.
by adding 3(_{1}H^{2}) → 2He^{4 }+ p + n
∴ Δm = 3(2.014)  [4.001 + 1.007 + 1.008] or Δm = 0.026 amu
Mass is converted into energy1 amu = 931.5 MeV or 1 amu = 931.5 x 10^{6} x 1.6 x 10^{19} J
∴ Energy from (Δm) = 0.026 x 931.5 x 1.6 x 10^{13 }J = 3.87 x 10^{12} J
∴ Energy released by 3 deuterons = 3.87 x 10^{12} J∴ Energy for 10^{40} deuterons
Time is of the order of 10^{12} sec.
Q.7. A nucleus X , initially at rest, undergoes alpha decay according to the equation,
(a) Find the values of A and Z in the above process.
(b) The alpha particle produced in the above process is found: to move in a circular track. of radius 0.11 m in a uniform magnetic field of 3 tesla. Find the energy (in MeV) released during the process and the binding energy of the parent nucleus X .
Given that:
(a) A = 228 + 4 = 232, Z = 92  2 = 90
(b) The magnetic force q vB provides centripetal force mv^{2}/r to α  particle for its circular motion.Linear momentum is conserved in the process.
m_{Y}v_{Y} = m_{α}v_{α} ....(ii)
Nucleus X is initially at rest
Total kinetic energy = K_{a} + K_{r}
Equivalent mass = 5.31/931.5 = 0.0057amu
X is parent nucleus, Y is daughter nucleusAlso m_{X} = m_{Y} + m_{α} + equivalent mass
= 228.03 + 4.003 + 0.0057 = 232.0387 amuNow, Nucleons = 92 protons + 140 neutron
Binding energy = (92 x 1.008) + (140 x 1.009)  232.0387
B.E. = 1.9573amu or B.E. = 1.9573 x 931.5 MeV
∴ B.E. of nucleus X = 1823MeV ...(iv)
Q.8. It is proposed to use the nuclear fusion reaction
in a nuclear reactor of 200 MW rating. If the energy from the above reaction is used with 25 percent efficiency in the reactor, how many gram of deuterium fuel will be needed per day. (The masses are 2.0141 atomic mass unit and 4.0026 atomic mass unit respectively).
Deuterium fuel needed per day in the reactor:
Mass defect provides the energy in the reactor:Δm = 2(2.0141)  (4.0026) = 4.0282  4.0026 = 0.0256amu
∴ ΔE (m)(931.5 MeV) (0.0256)(931.5)x1.6x10^{13} J = 3.82x10^{12} J
Efficiency = 25%
∴ Energy available due to fusion of two deuterium nuclei∴ Energy available = 9.55 x 10^{13} J
Energy available per deuterium nuclei
Total energy needed = power x time = (200 x 10^{6}) x (24 x 60 x 60) = 1.728x10^{13} J
Number of deuterium nuclei required∴ Mass of deuterium required = m
or Mass = 120.26gram
Hence 120.26 gram of deuterium fuel will be needed per day in the nuc1ear reactor
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