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Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

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Q.1. A nucleus at rest undergoes a decay emitting an α - particle of de-Broglie 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.
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

Let ZX4 represent the parent nucleus. The parent nucleus decays emitting α - particles
ZX4Z-2 YA-4 + 2He4

Mass of daughter nucleus Y = 223.610 amu , Mass of α - particle = 4.002 amu 

De-Broglie wavelength of α - particle, λ = 5.76 x 10-15 m
So, momentum of α - particle would be
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

From law of conservation of linear momentum, this should also be equal to the linear momentum the daughter nucleus thus pα = pY
Let Kα and KY be the kinetic energies of α - particle and daughter nucleus. Then total kinetic energy in the final state is
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

∵ 1 amu 1.67x10-27 kg   

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Mass equivalent to energy 

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
or Δm = 0.0067amu ∴ mX = mY + mα + Δm
or mX = (223.610 + 4.002 + 0.0067)amu or mX = 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.
∴ p1 = p2 
but Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics where K = kinetic energy
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
where K2 is for α - particle and K1 is for nucleus.
or 216K1 = 4K2 or K2 = 54K1 ....(i)
Given: K1 + K2 = 5.5MeV   .....(ii) 

From (i) and (ii), 

K1 + 54K1 = 5.5 (MeV)
or 55K1 = 5.5 (MeV) 

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
∴ K254 x K1 or Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics or K2 = 5.4 MeV    ....(iv)
∴ Kinetic energy of α - particle = 5.4 MeV  


Q.3. In a nuclear reaction 235U 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. 

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
∴ Equivalent energy = Power input x time = 1010 x 365 x 24 x 60 x 60 J
= 31.1536 x 1018 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),
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

or Number of fissions = 9.85 x 1028 

∴ Number of U235 atoms = 9.85 x 1028 
Avogadro number 6.02 x 1026 per kg 26

∴ 6.02 x 106 atoms are contained in 235kg of U235 

∴ 9.85 x 1028 atoms are contained in mass
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
∴ Mass of U235 = 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[2He4] →8 O16  

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 Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics has a mean life of 1013 second, its primary decay modes are spontaneous fission and α-decay, 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 α-decay are as follows:
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Calculate the power output from a sample of 1020Cm atoms.

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

The primary decay modes of 96Cm248 are

(i) Spontaneous fission (probability 8%)
(ii) - α decay (probability 92%).

The nuclear reaction is given as:  
96Cm24894Pu244 + 2He4  

∴ 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 

Eα = 5.136 MeV
Given: = Ef Each fission released 200 meV of energy

Mean life of 1013 sec 

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

again dN/dt = λN where N is given to be 1020

∴ Rate of decay = (10-13)(1020) or Rate of decay 107 dps 

of these dps 107 dps, 8% are in fission and 92% are in α - decay process.
Energy released per second due to fission
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Energy released per sec due to α - decay
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Total energy released per second = (16 + 4.725)107 MeV = 20.725 x 107 MeV
∴ Power output = Energy per second
= (20.725 x 107) 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 1040 deuterons. It produces energy via the processes 
1H2 + 1H21H3 + p, and 1H2 + 1H22 He4 + n.
If the average power radiated by the star is 1016W , then the time required for the deuteron supply of the star to be exhausted.
The masses of the nuclei are as follow:
M(H2) = 2.014 amu ; M(p) = 1.007 amu; M (n) = 1.008 amu; M (He4) = 4.001amu.

1H2 + 1H21H3 + p, and 1H2 + 1H22 He4 + n.

by adding 3(1H2) → 2He+ p + n
∴ Δm = 3(2.014) - [4.001 + 1.007 + 1.008] or Δm = 0.026 amu
Mass is converted into energy 

1 amu = 931.5 MeV or 1 amu = 931.5 x 106 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 1040 deuterons

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

Time is of the order of 1012 sec.


Q.7. A nucleus X , initially at rest, undergoes alpha decay according to the equation,
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
(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:
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

(a) A = 228 + 4 = 232, Z = 92 - 2 = 90
(b) The magnetic force q vB provides centripetal force mv2/r to α - particle for its circular motion.
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

Linear momentum is conserved in the process.
mYvY = mαvα ....(ii)
Nucleus X is initially at rest
Total kinetic energy = Ka + Kr   

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Equivalent mass = 5.31/931.5 = 0.0057amu
X is parent nucleus, Y is daughter nucleus

Also mX = mY + mα + equivalent mass
= 228.03 + 4.003 + 0.0057 = 232.0387 amu 

Now, 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
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

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 Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics 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 

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

∴ Energy available = 9.55 x 10-13 J
Energy available per deuterium nuclei

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
Total energy needed = power x time = (200 x 106) x (24 x 60 x 60) = 1.728x1013 J
Number of deuterium nuclei required
Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

∴ Mass of deuterium required = m

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics

Nuclear Reaction: Assignment Notes | Study Modern Physics - Physics
or Mass = 120.26gram
Hence 120.26 gram of deuterium fuel will be needed per day in the nuc1ear reactor

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