Nuclear Reaction: Assignment | Modern Physics PDF Download

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.

Let _ZX⁴ represent the parent nucleus. The parent nucleus decays emitting α - particles
∴ _ZX⁴ → _Z-2 Y^A-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

From 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.
∴ p₁ = p₂ 
but  where K = kinetic energy

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) 

∴ K₂54 x K₁ or  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. 

∴ 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),

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

∴ 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  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:

Calculate the power output from a sample of 10²⁰Cm atoms.

The primary decay modes of ₉₆Cm²⁴⁸ are

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

The nuclear reaction is given as:  
₉₆Cm²⁴⁸ → ₉₄Pu²⁴⁴ + 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 

E_α = 5.136 MeV
Given: = E_f Each fission released 200 meV of energy

Mean life of 10¹³ sec 

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

Energy released per sec due to α - decay

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 + ₁H² → ₁H³ + p, and ₁H² + ₁H² →₂ 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 + ₁H² → ₁H³ + p, and ₁H² + ₁H² →₂ He⁴ + n.

by adding 3(₁H²) → 2He⁴ + 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

Time is of the order of 10¹² 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 mv²/r to α - particle for its circular motion.

Linear momentum is conserved in the process.
m_Yv_Y = m_αv_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 nucleus

Also m_X = m_Y + m_α + 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

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 10⁶) x (24 x 60 x 60) = 1.728x10¹³ 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