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Question #43834

Difficulty: 8/10 | 🟢 Free

The Question

In a plasma fusion reactor, the fusion consumption rate per unit volume is modeled as Φ(n)=Anexp⁡(−nc/n)\Phi(n) = A n \exp(-n_c / n), where nn is the local particle density, nc=1.0×1020 m−3n_c = 1.0 \times 10^{20} \text{ m}^{-3} is a critical density parameter, and A=3.0e s−1A = 3.0 e \text{ s}^{-1} (ee being the base of the natural logarithm). The reactor is maintained at a steady-state baseline density n0=ncn_0 = n_c by a constant fuel injection. To probe the reactor's dynamic response, a small oscillating fuel injection rate ΔS(t)=Smcos⁡(ωt)\Delta S(t) = S_m \cos(\omega t) is superimposed, where Sm=1.0×1018 m−3s−1S_m = 1.0 \times 10^{18} \text{ m}^{-3}\text{s}^{-1} and ω=8.0 rad s−1\omega = 8.0 \text{ rad s}^{-1}. Assuming the density fluctuations Δn(t)\Delta n(t) remain small (Δn≪n0\Delta n \ll n_0), what is the amplitude of the resulting steady-state fluctuation in the fusion consumption rate, ΔΦmax\Delta \Phi_{\text{max}}?

Options

  • A.

    6.0×1017 m−3s−16.0 \times 10^{17} \text{ m}^{-3}\text{s}^{-1}

  • B.

    1.0×1018 m−3s−11.0 \times 10^{18} \text{ m}^{-3}\text{s}^{-1}

  • C.

    7.5×1017 m−3s−17.5 \times 10^{17} \text{ m}^{-3}\text{s}^{-1}

  • D.

    3.073×1018 m−3s−1\frac{3.0}{\sqrt{73}} \times 10^{18} \text{ m}^{-3}\text{s}^{-1}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Fusion Barriers|Radioactive Decay Law Dynamics|Dynamic Resistance (AC Resistance)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply dynamic resistance concepts to a non-linear statistical decay process.

The primary trap lies in misapplying standard linear decay laws (λ=constant\lambda = \text{constant}) to a system where the decay probability depends on density.

Students often incorrectly use the static average decay constant λavg=Φ(n0)/n0=3.0 s−1\lambda_{\text{avg}} = \Phi(n_0)/n_0 = 3.0 \text{ s}^{-1} instead of the dynamic decay constant γ=Φ′(n0)=6.0 s−1\gamma = \Phi'(n_0) = 6.0 \text{ s}^{-1}, analogous to confusing static resistance (V/IV/I) with dynamic resistance (dV/dIdV/dI) in a diode.

Additionally, treating the system as purely capacitive (ignoring consumption feedback) or purely resistive (ignoring the phase shift from the time derivative) leads to incorrect amplitudes.

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Question Information

SubjectPhysics
ChapterNuclei - Class 12
ConceptFusion Barriers
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree