At 2000 K , the saturation current from a thoriated-tungsten cathode is 100 mA . Determine the saturation current for a pure-tungsten cathode that has the same surface area and operates at the same temperature. In the

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 33 \,

According to the Richardson-Dushman equation, the saturation current is given by:

I = S A T^2 e^{-\Phi / (kT)}

For pure tungsten, the saturation current is:

I_p = S A_p T^2 e^{-\Phi_p / (kT)}

For thoriated tungsten, the saturation current is:

I_t = S A_t T^2 e^{-\Phi_t / (kT)}

Taking the ratio of the two currents:

\dfrac{I_p}{I_t} = \dfrac{A_p}{A_t} e^{-(\Phi_p - \Phi_t) / (kT)}

Substituting the given values:

\dfrac{A_p}{A_t} = \dfrac{60 \times 10^4}{3.0 \times 10^4} = 20

\Phi_p - \Phi_t = 4.5 \text{ eV} - 2.6 \text{ eV} = 1.9 \text{ eV}

kT = \dfrac{1.38 \times 10^{-23} \times 2000}{1.6 \times 10^{-19}} \text{ eV} = 0.1725 \text{ eV}

Calculating the exponent term:

\dfrac{\Phi_p - \Phi_t}{kT} = \dfrac{1.9}{0.1725} \approx 11.01

e^{-11.01} \approx 1.65 \times 10^{-5}

Therefore, the ratio becomes:

\dfrac{I_p}{I_t} = 20 \times 1.65 \times 10^{-5} = 3.3 \times 10^{-4}

Since

I_t = 100 \text{ mA}

:

I_p = 100 \text{ mA} \times 3.3 \times 10^{-4} = 33 \times 10^{-3} \text{ mA} = 33 \, \mu\text{A}

At 2000 K , the saturation current from a thoriated-tungsten...

Options

Correct answer

Step-by-step solution

More from Electric Current through Gases