A pure tungsten cathode and a thoriated-tungsten cathode share the same geometrical dimensions and are operated at a common temperature. The current produced by the thoriated-tungsten cathode is 5000 times greater than

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 1914 K

Using the Richardson-Dushman equation, the emission current is given by:

I = S A T^2 e^{-\dfrac{\Phi}{kT}}

For pure tungsten:

I_1 = S A_1 T^2 e^{-\dfrac{\Phi_1}{kT}}

For thoriated tungsten:

I_2 = S A_2 T^2 e^{-\dfrac{\Phi_2}{kT}}

Given that

I_2 = 5000 I_1

, we get:

S A_2 T^2 e^{-\dfrac{\Phi_2}{kT}} = 5000 S A_1 T^2 e^{-\dfrac{\Phi_1}{kT}}

\dfrac{A_2}{A_1} e^{\dfrac{\Phi_1 - \Phi_2}{kT}} = 5000

Substituting the given values

A_1 = 60 \times 10^4 \text{ A m}^{-2} \text{ K}^{-2}

,

A_2 = 3.0 \times 10^4 \text{ A m}^{-2} \text{ K}^{-2}

,

\Phi_1 = 4.5 \text{ eV}

, and

\Phi_2 = 2.6 \text{ eV}

:

\dfrac{3.0 \times 10^4}{60 \times 10^4} e^{\dfrac{4.5 - 2.6}{kT}} = 5000

\dfrac{1}{20} e^{\dfrac{1.9 \text{ eV}}{kT}} = 5000

e^{\dfrac{1.9 \text{ eV}}{kT}} = 100000 = 10^5

Taking the natural logarithm on both sides:

\dfrac{1.9 \text{ eV}}{kT} = 5 \ln 10

Using the standard value

\dfrac{1 \text{ eV}}{k} \approx 11600 \text{ K}

and

\ln 10 \approx 2.303

:

T = \dfrac{1.9 \times 11600}{5 \times 2.303}

T = \dfrac{22040}{11.515} \approx 1914 \text{ K}

A pure tungsten cathode and a thoriated-tungsten cathode share the...

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