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Q1. (a) Distinguish between Galilean and Einsteinian principles of relativity. (b) The mean life time of a free neutron at rest is measured to be 900 s. What is its mean life time measured by an observer with respect to whom the neutron travels at a speed of 0.9 c ? (c) Calculate the de Broglie’s wavelength of a 1.0 MeV electron. (d) What is the probabilistic interpretation of the wave function ? (e) Define and draw a step potential of step V₀>0 at x=0. (f) Explain briefly how quantum tunneling enables alpha decay from the atomic nucleus. (g) What is secular equilibrium in radioactivity ? (h) Write the functions of shim rods and safety rods in a nuclear reactor.

Ans. (a) Distinction between Galilean and Einsteinian Principles of Relativity: Both Galilean and Einsteinian principles of relativity describe the laws of physics in different reference frames, but they are based on fundamentally different postulates.

• Galilean Principle of Relativity:
  • It states that the laws of mechanics are the same in all inertial frames of reference.
  • It assumes that time is absolute , meaning it flows at the same rate for all observers, regardless of their motion.
  • Velocities add up according to Galilean transformation (e.g., u’ = u – v).
• Einsteinian Principle of Relativity (Special Relativity):
  • It extends the principle to all laws of physics (not just mechanics), stating they are the same in all inertial frames.
  • Its second postulate is the constancy of the speed of light . The speed of light in a vacuum (c) is the same for all inertial observers, irrespective of the motion of the source or the observer.
  • It deduces that space and time are not absolute ; they are relative and depend on the observer’s motion (leading to phenomena like time dilation and length contraction).

The key difference is Einstein’s inclusion of the constancy of the speed of light, which revolutionizes the classical concepts of space and time, whereas Galileo’s principle is a low-velocity approximation and treats time as absolute.

(b) This problem involves the concept of time dilation . According to special relativity, a moving clock runs slower as measured by a stationary observer. Given: Proper mean lifetime of the neutron (at rest), t₀ = 900 s Speed of the neutron, v = 0.9 c The formula for time dilation is: t = γt₀ = t₀ / √(1 – v²/c²) where t is the lifetime measured by the observer. Substituting the values: t = 900 / √(1 – (0.9c)²/c²) t = 900 / √(1 – 0.81) t = 900 / √0.19 t = 900 / 0.4359 t ≈ 2064.7 s Thus, the mean lifetime of the moving neutron as measured by the observer will be approximately 2064.7 seconds .

(c) The kinetic energy (K) of the electron is 1.0 MeV. The rest energy of an electron (E₀ = m₀c²) is approximately 0.511 MeV. Since the given kinetic energy is comparable to the rest energy, we must use relativistic formulas. The total energy (E) of the electron is: E = K + E₀ = 1.0 MeV + 0.511 MeV = 1.511 MeV From the energy-momentum relation: E² = (pc)² + (E₀)² where p is the momentum. (pc)² = E² – E₀² (pc)² = (1.511)² – (0.511)² MeV² (pc)² = 2.283 – 0.261 MeV² = 2.022 MeV² pc = √2.022 MeV ≈ 1.422 MeV Now, the de Broglie wavelength (λ) is given by λ = h/p . We can rewrite this as λ = hc/pc . h = 6.63 × 10⁻³⁴ Js, c = 3 × 10⁸ m/s It is convenient to use the value hc ≈ 1240 eV·nm = 1240 × 10⁻³ MeV·nm. pc = 1.422 MeV λ = (1240 × 10⁻³ MeV·nm) / (1.422 MeV) λ ≈ 0.872 × 10⁻³ nm λ ≈ 8.72 × 10⁻¹³ m So, the de Broglie wavelength of the 1.0 MeV electron is approximately 8.72 × 10⁻¹³ meters .

(d) In quantum mechanics, the state of a particle is described by a wave function, Ψ(x, t) . The probabilistic interpretation of the wave function was proposed by Max Born. According to this interpretation:

• The wave function Ψ(x, t) itself is not a physically measurable quantity. It can be a complex quantity.
• The quantity |Ψ(x, t)|² = Ψ (x, t)Ψ(x, t) , where Ψ is the complex conjugate of Ψ, is called the probability density .
• This probability density gives a measure of the probability of finding the particle between position x and x + dx at time t. That is, P(x, dx) = |Ψ(x, t)|² dx .
• The total probability of finding the particle in a certain region is obtained by integrating the probability density over that region. The total probability of finding the particle anywhere in space must be 1, which is the normalization condition: ∫|Ψ(x, t)|² dx = 1 (over all space).

In short, |Ψ|² tells us how likely it is to find the particle at a given location.

(e) Step Potential: A step potential is a theoretical potential that changes abruptly at a single point. It is mathematically defined as: V(x) = { 0, for x < 0 V(x) = { V₀, for x ≥ 0 where V₀ > 0 is a positive constant, called the height of the step. This means a particle feels no force (zero potential) in the region x < 0, but as soon as it crosses x = 0, it experiences a constant repulsive potential V₀. Drawing: The graph of the step potential plots position (x) on the x-axis and potential (V(x)) on the y-axis. In this figure, the potential is zero for the negative part of the x-axis, and at x=0, it abruptly rises to a height V₀ and remains constant for all positive values of x.

(f) Alpha Decay and Quantum Tunneling: Alpha decay is the process where a heavy, unstable nucleus decays by emitting an alpha particle (a helium nucleus).

• Inside the nucleus, the alpha particle is attracted by the strong nuclear force, but once it is just outside the nucleus, it is strongly repelled by the Coulomb force from the positive charge of the remaining nucleus.
• This situation creates a potential barrier . The alpha particle is trapped inside this barrier.
• According to classical physics, if the alpha particle’s energy is less than the height of the potential barrier, it can never escape the nucleus.
• According to quantum mechanics, the particle is described by a wave function. This wave function is not zero inside the barrier but penetrates into it as a decaying exponential function.
• Because the wave function is non-zero on the other side of the barrier as well, there is a non-zero probability that the alpha particle can “tunnel” through the barrier and escape the nucleus, even though its energy is less than the barrier height.

Thus, the phenomenon of quantum tunneling explains alpha decay, which is impossible according to classical physics.

(g) Secular Equilibrium: Secular equilibrium is a special type of radioactive equilibrium that occurs in a decay chain when the half-life of the parent nucleus (T₁) is much, much longer than the half-life of the daughter nucleus (T₂) (i.e., T₁ >> T₂) . In this situation:

• Initially, the number of daughter nuclei builds up as the parent decays, and so the activity of the daughter increases.
• After a certain time (several daughter half-lives), the rate at which the daughter is decaying becomes almost equal to the rate at which it is being produced by the parent.
• At this point, the activity of the daughter (A₂) becomes nearly equal to the activity of the parent (A₁): A₁ ≈ A₂ .
• Because the parent’s half-life is so long, its activity remains almost constant over the timescale of the experiment. Therefore, the daughter’s activity also appears constant, tracking the decay rate of the parent. The ratio of their activities becomes constant.

A classic example is the equilibrium between Uranium-238 (T₁ ≈ 4.5 billion years) and Thorium-234 (T₂ ≈ 24.1 days).

(h) In a nuclear reactor, shim rods and safety rods are both made of neutron-absorbing material (like cadmium or boron) and are used to control the reaction rate, but they have distinct functions.

• Shim Rods:
  • These are used for slow, coarse control of the reactor’s power level.
  • They are designed to compensate for long-term changes in reactivity during operation, such as fuel burn-up or the build-up of fission product poisons like Xenon.
  • They are also adjusted slowly during the process of starting up or shutting down the reactor.
• Safety Rods:
  • Their primary function is to provide for a rapid emergency shutdown of the reactor, a procedure known as a SCRAM .
  • These rods are held completely out of the reactor core during normal operation.
  • Upon any sign of an abnormal or dangerous condition (like a sudden rise in temperature or pressure), these rods are instantly dropped or driven into the core by gravity or another fast-acting mechanism.
  • By inserting a large amount of neutron-absorbing material, they absorb a massive number of neutrons, shutting down the chain reaction almost immediately and making the reactor safe.

Q2. (a) Show that simultaneous events that occur in an inertial frame of reference S at different positions are not simultaneous in another inertial frame of reference S’ moving uniformly with respect to S. (b) A spaceship is receding from the Earth at a speed of 0.25 c. An observer in the spaceship measures the wavelength of light from a source in it as 500 nm. What is the wavelength of this light as measured by an observer on the Earth ? (c) The rest energy of a particle is 1000 MeV. Calculate its linear momentum and total energy (in MeV) given that it travels at a speed of 0.6 c.

Ans. (a) Relativity of Simultaneity We want to prove that two events that are simultaneous in one inertial frame S, but occur at different locations, will not be simultaneous in another inertial frame S’ moving at a constant velocity v with respect to S. Let’s consider two events (Event 1 and Event 2) in frame S. Coordinates of Event 1 in S: (x₁, t₁) Coordinates of Event 2 in S: (x₂, t₂)

According to the problem, the events are simultaneous in S, which means they occur at the same time: t₁ = t₂ And they occur at different positions: x₁ ≠ x₂

Now, let’s find the times of these events in frame S’, which is moving with velocity v along the x-axis relative to S. We use the Lorentz transformation equation for time: t’ = γ (t – vx/c²) where γ = 1 / √(1 – v²/c²)

The time of Event 1 in frame S’ will be: t’₁ = γ (t₁ – vx₁/c²) The time of Event 2 in frame S’ will be: t’₂ = γ (t₂ – vx₂/c²)

The time difference between these events in S’ is: Δt’ = t’₂ – t’₁ Δt’ = γ (t₂ – vx₂/c²) – γ (t₁ – vx₁/c²) Δt’ = γ [(t₂ – t₁) – (v/c²)(x₂ – x₁)]

Since the events are simultaneous in S, t₁ = t₂, so (t₂ – t₁) = 0. Therefore, Δt’ = – (γv/c²) (x₂ – x₁)

Because v ≠ 0 and we have assumed that x₁ ≠ x₂, it means that (x₂ – x₁) ≠ 0. Also, γ ≠ 0. Thus, Δt’ ≠ 0 .

This means that t’₁ ≠ t’₂, and the two events do not occur at the same time in S’. They are not simultaneous in S’. This proves that simultaneity is not absolute but depends on the observer’s state of motion.

(b) This problem relates to the Relativistic Doppler Effect . When the source of light is moving away from the observer, the observed wavelength (λ_obs) is longer than the emitted wavelength (λ_src), a phenomenon known as redshift. Given: Speed of the spaceship, v = 0.25 c (receding from Earth) Wavelength of light emitted by the source (measured in the spaceship), λ_src = 500 nm We will use β = v/c = 0.25.

The formula for the relativistic Doppler effect when the source is receding from the observer is: λ_obs = λ_src * √[(1 + β) / (1 – β)]

Substituting the values: λ_obs = 500 nm * √[(1 + 0.25) / (1 – 0.25)] λ_obs = 500 nm * √[1.25 / 0.75] λ_obs = 500 nm * √[5 / 3] λ_obs = 500 nm * √1.6667 λ_obs = 500 nm * 1.291 λ_obs ≈ 645.5 nm

Therefore, the wavelength of the light as measured by an observer on Earth would be approximately 645.5 nm . This value is longer than the emitted wavelength, consistent with the expected redshift.

(c) For the given particle: Rest energy, E₀ = 1000 MeV Speed of the particle, v = 0.6 c

First, let’s calculate the Lorentz factor (γ): γ = 1 / √(1 – v²/c²) γ = 1 / √(1 – (0.6c)²/c²) γ = 1 / √(1 – 0.36) γ = 1 / √0.64 γ = 1 / 0.8 γ = 1.25

1. Total Energy: The total relativistic energy (E) of the particle is related to its rest energy (E₀) and the Lorentz factor (γ) by: E = γE₀ E = 1.25 × 1000 MeV E = 1250 MeV

So, the total energy of the particle is 1250 MeV .

2. Linear Momentum: We can calculate the relativistic momentum (p) using the energy-momentum relation: E² = (pc)² + E₀² Solving for p: (pc)² = E² – E₀² (pc)² = (1250 MeV)² – (1000 MeV)² (pc)² = 1,562,500 – 1,000,000 (pc)² = 562,500 (MeV)² pc = √562,500 MeV pc = 750 MeV

Momentum is often expressed in units of energy divided by c (MeV/c). Therefore, the linear momentum is p = 750 MeV/c .

Final answer:

• Total Energy = 1250 MeV
• Linear Momentum = 750 MeV/c

Q3. (a) The wave function of a particle of mass m and zero energy is given by : Ψ(x) = Ne⁻²ˣ, where N is a constant. Write down the time-independent Schrödinger equation for the particle and determine its potential energy. (b) Use the uncertainty principle to explain the concept of zero-point energy. (c) Calculate ⟨x⟩ for the wave function Ψ(x) = √(2/L) cos(πx/L), 0 < x < L.

Ans. (a) Schrödinger Equation and Potential Energy Given: Mass of the particle = m Total energy of the particle, E = 0 Wave function, Ψ(x) = Ne⁻²ˣ

The one-dimensional Time-Independent Schrödinger Equation (TISE) is: (-ħ²/2m) d²Ψ/dx² + V(x)Ψ(x) = EΨ(x)

First, we need to differentiate the given wave function twice with respect to x: First derivative: dΨ/dx = d/dx (Ne⁻²ˣ) = -2Ne⁻²ˣ

Second derivative: d²Ψ/dx² = d/dx (-2Ne⁻²ˣ) = (-2)(-2)Ne⁻²ˣ = 4Ne⁻²ˣ

Since Ψ(x) = Ne⁻²ˣ, we can write: d²Ψ/dx² = 4Ψ(x)

Now substitute these values (E=0 and the second derivative) into the TISE: (-ħ²/2m) (4Ψ(x)) + V(x)Ψ(x) = (0)Ψ(x) -2ħ²Ψ(x)/m + V(x)Ψ(x) = 0

Dividing through by Ψ(x) (assuming Ψ(x) is non-zero): -2ħ²/m + V(x) = 0

Solving for V(x): V(x) = 2ħ²/m

This result shows that the potential energy V(x) does not depend on position x; it is a constant potential . Therefore, the potential energy of the particle is V = 2ħ²/m .

(b) Uncertainty Principle and Zero-Point Energy Zero-Point Energy is the lowest possible energy that a quantum mechanical system may have, even at absolute zero temperature (0 K). This concept arises directly from the Heisenberg Uncertainty Principle.

The Uncertainty Principle states that the product of the uncertainties in a particle’s position (Δx) and momentum (Δp) must be greater than or equal to a certain minimum value: Δx Δp ≥ ħ/2

Now, consider a particle confined within a potential well, such as a box of length L.

1. Uncertainty in Position: Since the particle is confined within the box, the maximum uncertainty in its position can be no more than the length of the box, i.e., Δx ≈ L .
2. Uncertainty in Momentum: Using the uncertainty principle, there must be a minimum uncertainty in the momentum: Δp ≥ ħ / (2Δx) ≈ ħ / (2L) This means the momentum of the particle cannot be precisely zero. If the momentum were zero, Δp would be zero, violating the uncertainty principle.
3. Minimum Kinetic Energy: The kinetic energy (K) of the particle is given by p²/2m. Since the particle has a non-zero momentum uncertainty (Δp), it means that while its average momentum could be zero, its average squared momentum (⟨p²⟩) will not be zero. The value of ⟨p²⟩ will be at least of the order of (Δp)². Therefore, the minimum average kinetic energy will be: K_min ≈ (Δp)² / 2m ≥ (ħ / (2L))² / 2m = ħ² / (8mL²) This energy is non-zero.

This minimum, non-zero energy is the

zero-point energy

. It is a purely quantum effect demonstrating that a confined particle can never be completely at rest; it must always possess some kinetic energy.

(c) Calculation of ⟨x⟩ The expectation value of position, ⟨x⟩, is given by the formula: ⟨x⟩ = ∫ Ψ*(x) x Ψ(x) dx The limits of integration are over the region where the particle exists, which is from 0 to L. The given wave function is: Ψ(x) = √(2/L) cos(πx/L) This is a real function, so Ψ*(x) = Ψ(x).

Substituting into the formula for ⟨x⟩: ⟨x⟩ = ∫₀ᴸ [√(2/L) cos(πx/L)] x [√(2/L) cos(πx/L)] dx ⟨x⟩ = (2/L) ∫₀ᴸ x cos²(πx/L) dx

To solve the integral, we use the trigonometric identity cos²(θ) = (1 + cos(2θ))/2 . Here, θ = πx/L, so cos²(πx/L) = (1 + cos(2πx/L))/2. ⟨x⟩ = (2/L) ∫₀ᴸ x [(1 + cos(2πx/L))/2] dx ⟨x⟩ = (1/L) ∫₀ᴸ [x + x cos(2πx/L)] dx ⟨x⟩ = (1/L) [∫₀ᴸ x dx + ∫₀ᴸ x cos(2πx/L) dx]

Now we solve the two integrals separately: First integral: ∫₀ᴸ x dx = [x²/2]₀ᴸ = L²/2

Second integral: (using integration by parts) ∫ u dv = uv – ∫ v du Let u = x (⇒ du = dx) and dv = cos(2πx/L) dx v = ∫ cos(2πx/L) dx = (L/2π) sin(2πx/L) ∫₀ᴸ x cos(2πx/L) dx = [x (L/2π) sin(2πx/L)]₀ᴸ – ∫₀ᴸ (L/2π) sin(2πx/L) dx = [ (L²/2π) sin(2π) – 0 ] – (L/2π) [(-L/2π) cos(2πx/L)]₀ᴸ = 0 – (L/2π) [-L/2π (cos(2π) – cos(0))] = (L²/4π²) [1 – 1] = 0

Plugging both results back into the equation for ⟨x⟩: ⟨x⟩ = (1/L) [L²/2 + 0] ⟨x⟩ = L/2

Alternatively: The probability density for the given wave function, |Ψ(x)|² = (2/L)cos²(πx/L), is symmetric about x=L/2. Whenever the probability density is symmetric about a point, the expectation value of the position is that point. Therefore, even without calculation, we could have deduced that ⟨x⟩ = L/2.

Q4. (a) Write down the time-independent Schrödinger equation for a particle of mass m in a one-dimensional potential barrier of height V₀ and width a. Obtain the general solutions of the equation when the energy of the particle E > V₀. Write down the boundary conditions for the problem. (b) Show that for a symmetric potential function [V(x) = V(-x)], the parity operator commutes with the Hamiltonian.

Ans. (a) Potential Barrier with E > V₀ A one-dimensional potential barrier is defined by the potential: V(x) = { V₀, for 0 < x < a V(x) = { 0, for x ≤ 0 (Region I) and x ≥ a (Region III) where V₀ is the height and a is the width of the barrier. We are considering the case where the particle’s energy E is greater than the barrier height (E > V₀).

1. Time-Independent Schrödinger Equation (TISE): The general TISE for the particle is: d²Ψ/dx² + (2m/ħ²) [E – V(x)] Ψ(x) = 0

We solve this equation in the three distinct regions:

• Region I (x < 0): Here, V(x) = 0. The TISE becomes: d²Ψ₁/dx² + (2mE/ħ²) Ψ₁ = 0 This can be written as d²Ψ₁/dx² + k₁² Ψ₁ = 0, where k₁ = √(2mE)/ħ . The general solution is: Ψ₁(x) = A e^(ik₁x) + B e^(-ik₁x) (A e^(ik₁x) represents the incident wave and B e^(-ik₁x) represents the reflected wave).
• Region II (0 < x < a): Here, V(x) = V₀. The TISE becomes: d²Ψ₂/dx² + (2m(E – V₀)/ħ²) Ψ₂ = 0 This can be written as d²Ψ₂/dx² + k₂² Ψ₂ = 0, where k₂ = √(2m(E – V₀))/ħ . (Since E > V₀, k₂ is real). The general solution is: Ψ₂(x) = C e^(ik₂x) + D e^(-ik₂x)
• Region III (x > a): Here, V(x) = 0, same as Region I. The TISE is: d²Ψ₃/dx² + (2mE/ħ²) Ψ₃ = 0 or d²Ψ₃/dx² + k₁² Ψ₃ = 0 The general solution is: Ψ₃(x) = F e^(ik₁x) + G e^(-ik₁x) Since there is no particle incident from the right in Region III, there can be no reflected wave. Thus, G = 0. Ψ₃(x) = F e^(ik₁x) (F e^(ik₁x) represents the transmitted wave).

2. Boundary Conditions: For the wave function to be physically acceptable, the wave function Ψ(x) and its first derivative dΨ/dx must be continuous everywhere, particularly at the boundaries where the potential changes (x=0 and x=a). This gives us the following four boundary conditions:

1. Continuity of Ψ at x = 0: Ψ₁(0) = Ψ₂(0) ⇒ A + B = C + D
2. Continuity of dΨ/dx at x = 0: dΨ₁/dx|ₓ=₀ = dΨ₂/dx|ₓ=₀ ⇒ ik₁(A – B) = ik₂(C – D)
3. Continuity of Ψ at x = a: Ψ₂(a) = Ψ₃(a) ⇒ C e^(ik₂a) + D e^(-ik₂a) = F e^(ik₁a)
4. Continuity of dΨ/dx at x = a: dΨ₂/dx|ₓ=ₐ = dΨ₃/dx|ₓ=ₐ ⇒ ik₂(C e^(ik₂a) – D e^(-ik₂a)) = ik₁F e^(ik₁a)

These four equations can be used to solve for the coefficients B, C, D, and F in terms of the incident amplitude A, which allows for the calculation of the reflection and transmission coefficients.

(b) Parity Operator and the Hamiltonian We need to show that for a symmetric potential function [V(x) = V(-x)], the Parity Operator (P) commutes with the Hamiltonian Operator (H). That is, we must prove: [P, H] = 0 , or PH = HP .

Definition of Operators:

• Hamiltonian Operator (H): For a one-dimensional system, H is the sum of the kinetic and potential energy operators. H = T + V = (-ħ²/2m) d²/dx² + V(x)
• Parity Operator (P): This operator acts on a function f(x) by replacing x with -x. P f(x) = f(-x)

Evaluating the Commutator [P, H]: To see if P and H commute, we apply their commutator [P, H] to an arbitrary test function Ψ(x): [P, H] Ψ(x) = (PH – HP) Ψ(x) = PHΨ(x) – HPΨ(x)

We evaluate the two terms separately:

First Term: PHΨ(x) First apply H to Ψ(x), then apply P to the result. HΨ(x) = (-ħ²/2m) d²Ψ(x)/dx² + V(x)Ψ(x) PHΨ(x) = P [(-ħ²/2m) d²Ψ(x)/dx² + V(x)Ψ(x)] Using the linearity of operator P: PHΨ(x) = (-ħ²/2m) P(d²Ψ(x)/dx²) + P(V(x)Ψ(x)) Now, apply P (change x to -x): P(d²Ψ(x)/dx²) = d²Ψ(-x)/d(-x)² = d²Ψ(-x)/dx² (since (-x)² = x²) P(V(x)Ψ(x)) = V(-x)Ψ(-x) So, PHΨ(x) = (-ħ²/2m) d²Ψ(-x)/dx² + V(-x)Ψ(-x)

Second Term: HPΨ(x) First apply P to Ψ(x), then apply H to the result. PΨ(x) = Ψ(-x) HPΨ(x) = H[Ψ(-x)] Now apply H to Ψ(-x): H[Ψ(-x)] = (-ħ²/2m) d²/dx² [Ψ(-x)] + V(x) [Ψ(-x)] HPΨ(x) = (-ħ²/2m) d²Ψ(-x)/dx² + V(x)Ψ(-x)

Comparison: Now we compare the expressions for PHΨ(x) and HPΨ(x). PHΨ(x) = (-ħ²/2m) d²Ψ(-x)/dx² + V(-x) Ψ(-x) HPΨ(x) = (-ħ²/2m) d²Ψ(-x)/dx² + V(x) Ψ(-x)

These two expressions will be equal if and only if V(-x) = V(x) . The problem states that the potential is symmetric, i.e., V(x) = V(-x). Under this condition, PHΨ(x) = HPΨ(x). Since this is true for any arbitrary function Ψ(x), we can conclude that the operators themselves are equal: PH = HP or, [P, H] = PH – HP = 0

Thus, for a symmetric potential, the parity operator commutes with the Hamiltonian. An important consequence of this is that the energy eigenfunctions in a symmetric potential can be chosen to be functions of definite parity (either even or odd).

Q5. (a) Define ‘activity’ of a radioactive substance. The half life of an element is 6.93 days. After how many days will only 10% of the element be left ? (b) Calculate the binding energy of a ⁴He nucleus using the semi-empirical mass formula. Take α = 15.8, β = 17.8, γ = 23.7, δ = 0.71 and ε = 34. (c) Explain how a self-sustained fission chain reaction is achieved in a nuclear reactor.

Ans. (a) Activity and Radioactive Decay Activity: The activity (A) of a radioactive sample is defined as the rate at which decays occur in the sample, i.e., the number of disintegrations per unit time. If a sample contains N radioactive nuclei at time t and the decay constant is λ, the activity is given by: A = -dN/dt = λN The SI unit of activity is the Becquerel (Bq) , where 1 Bq = 1 decay per second. Another common unit is the Curie (Ci), where 1 Ci = 3.7 × 10¹⁰ Bq.

Calculation: Given: Half-life, T₁/₂ = 6.93 days We need to find the time (t) after which 10% of the element is left. This means N(t) / N₀ = 10/100 = 0.1, where N₀ is the initial number of nuclei and N(t) is the number of nuclei remaining at time t.

First, we calculate the decay constant (λ): λ = 0.693 / T₁/₂ λ = 0.693 / 6.93 days λ = 0.1 day⁻¹

Now, use the law of radioactive decay: N(t) = N₀ e^(-λt) N(t) / N₀ = e^(-λt) 0.1 = e^(-0.1 * t)

Take the natural logarithm of both sides to solve for t: ln(0.1) = -0.1 * t We know that ln(0.1) = ln(1/10) = -ln(10) ≈ -2.303. -2.303 = -0.1 * t t = -2.303 / -0.1 t = 23.03 days

Therefore, after 23.03 days , only 10% of the element will be left.

(b) Binding Energy from Semi-Empirical Mass Formula (SEMF) The Semi-Empirical Mass Formula (SEMF) estimates the binding energy (B) of a nucleus: B(A, Z) = a_vA – a_sA²/³ – a_c Z(Z-1)/A¹/³ – a_a (A-2Z)²/A + δ_p Matching the given coefficients to the standard terms (assuming α=a_v, β=a_s, δ=a_c, γ=a_a, ε=a_p):

• Volume term coefficient, a_v (α) = 15.8 MeV
• Surface term coefficient, a_s (β) = 17.8 MeV
• Coulomb term coefficient, a_c (δ) = 0.71 MeV
• Asymmetry term coefficient, a_a (γ) = 23.7 MeV
• Pairing term coefficient, a_p (ε) = 34 MeV

We need to calculate the binding energy for the ⁴He nucleus (alpha particle):

Mass number,

A = 4

Atomic number,

Z = 2

Neutron number, N = A – Z = 2

Now we calculate each term: 1. Volume Energy: E_v = a_v * A = 15.8 × 4 = 63.2 MeV 2. Surface Energy: E_s = -a_s * A²/³ = -17.8 × (4)²/³ = -17.8 × 2.5198 ≈ -44.85 MeV 3. Coulomb Energy: E_c = -a_c * Z(Z-1)/A¹/³ = -0.71 × 2(2-1)/(4)¹/³ = -0.71 × 2 / 1.5874 ≈ -0.89 MeV 4. Asymmetry Energy: E_a = -a_a * (A-2Z)²/A = -23.7 × (4 – 2×2)² / 4 = -23.7 × (0)² / 4 = 0 MeV 5. Pairing Energy: ⁴He is an even-even nucleus (Z=2, N=2). The pairing term is positive for even-even nuclei. A common form is `+a_p/A^(3/4)` which works well with the given constants. δ_p = +a_p / A³/⁴ = +34 / (4)³/⁴ = +34 / (2²)³/⁴ = +34 / 2¹·⁵ = +34 / 2.828 ≈ +12.02 MeV

Total Binding Energy (B): B = E_v + E_s + E_c + E_a + δ_p B = 63.2 – 44.85 – 0.89 + 0 + 12.02 B ≈ 29.48 MeV

Therefore, the binding energy of the ⁴He nucleus calculated using the SEMF is approximately 29.5 MeV . (This is quite close to the experimental value of 28.3 MeV).

(c) Self-Sustained Fission Chain Reaction A self-sustained fission chain reaction in a nuclear reactor means establishing a process where the rate of fission remains constant, leading to a controlled and continuous release of energy. This is achieved through the following key components and principles:

1. Fission: The process starts when a fissile nucleus, like Uranium-235 (²³⁵U) , absorbs a neutron. This excites it into an unstable state, and it splits into two or more smaller nuclei (fission products), releasing a large amount of energy and, on average, 2 to 3 new neutrons.

2. Chain Reaction: The neutrons produced from one fission event can go on to induce fission in other ²³⁵U nuclei. If, on average, at least one of these new neutrons causes another fission event, the reaction will continue on its own.

3. Neutron Multiplication Factor (k): The state of a chain reaction is described by the multiplication factor (k) : k = (number of fissions in one generation) / (number of fissions in the preceding generation)

• k < 1 (Subcritical): Fewer neutrons cause fissions in each generation. The reaction dies out.
• k > 1 (Supercritical): More neutrons cause fissions in each generation. The reaction rate grows exponentially (the principle of an atomic bomb).
• k = 1 (Critical): The reaction rate is stable. Exactly one neutron from each fission, on average, causes another fission. This is the desired self-sustained state in a reactor.

4. Achieving k=1:

• Moderator: Neutrons from fission are very fast (high energy). ²³⁵U is much more effective at capturing thermal (slow) neutrons. A moderator (like heavy water, ordinary water, or graphite) slows down the neutrons by colliding with them, making them more likely to cause fission.
• Control Rods: These are made of neutron-absorbing materials like boron or cadmium. By moving them into or out of the reactor core, excess neutrons can be absorbed, allowing the value of k to be precisely adjusted and maintained at 1.
• Critical Mass: A sufficient amount of fissile material (like ²³⁵U) is needed to sustain the chain reaction, so that the rate of fission is greater than the rate of neutron loss from the core’s surface.

By carefully balancing these mechanisms, a nuclear reactor achieves a stable, self-sustained chain reaction, producing a continuous output of energy at a controlled rate.