Thermionic emission – effect of a small temperature rise and Richardson–Dushman law Assertion (A): Raising the cathode temperature from 2500 K to 2550 K can increase thermionic emission current by roughly 50%. Reason (R): The emission current density obeys J ∝ T^2 * exp(−phi/(k T)).
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ABoth A and R are true and R is the correct explanation of A
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BBoth A and R are true but R is not the correct explanation of A
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CA is true but R is false
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DA is false but R is true
Answer
Correct Answer: Both A and R are true and R is the correct explanation of A
Explanation
Introduction / Context:Thermionic emission is the process by which electrons escape from a heated metal or oxide cathode. The emitted current density depends sensitively on temperature through an exponential factor, which explains why even a small temperature rise can yield a large current increase in vacuum tubes and electron guns.
Given Data / Assumptions:
- Initial temperature T1 = 2500 K, final temperature T2 = 2550 K.
- Richardson–Dushman relation: J = A T^2 * exp(−phi/(k T)), where phi is work function and k is Boltzmann constant.
- Typical metallic work functions are around 4 to 5 eV.
Concept / Approach:
The T^2 prefactor and the strong exponential term with 1/T together control sensitivity. A small increase in T reduces the exponent magnitude and increases the T^2 term, producing a substantial rise in J. This quantitatively supports the assertion of a roughly 50% increase for a 50 K rise at around 2500 K.
Step-by-Step Solution:
Use ratio J2/J1 = (T2/T1)^2 * exp[−phi/k * (1/T2 − 1/T1)].Compute (T2/T1)^2 = (2550/2500)^2 ≈ 1.040.Take phi ≈ 4.5 eV, k ≈ 8.617×10^−5 eV/K. Then 1/T2 − 1/T1 ≈ −7.84×10^−6 K^−1.Exponent term ≈ exp( (4.5/8.617×10^−5) * 7.84×10^−6 ) ≈ exp(0.41) ≈ 1.51.Therefore J2/J1 ≈ 1.040 * 1.51 ≈ 1.57, i.e., about 57% increase, consistent with the stated 50%.Verification / Alternative check:
Changing phi between 4 and 5 eV still yields a change on the order of 40–70%, validating the claim's order of magnitude.
Why Other Options Are Wrong:
Since the numerical estimate aligns with the law, both A and R are true and R directly explains A; other combinations do not match the physics.
Common Pitfalls:
Ignoring the exponential sensitivity, or assuming linear behavior with temperature. Also, confusing cathode temperature limits with emission saturation due to space charge, which is a separate effect.
Final Answer:
Both A and R are true and R is the correct explanation of A