For tungsten, the constant A in the Richardson–Dushman equation is 60 imes 10^4 A m^ -2 K^ -2 , and its work function is 4.5 eV . An electric heater of 24 W is used to heat a tungsten cathode with a surface area of 2.0

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 0.001 A

In the steady state, the power supplied by the heater is equal to the power radiated by the cathode as a blackbody.

P = \sigma S T^4

Substituting the given values:

24 = (6 \times 10^{-8}) \times (2.0 \times 10^{-5}) \times T^4

24 = 12 \times 10^{-13} \times T^4

T^4 = \dfrac{24}{12 \times 10^{-13}} = 2 \times 10^{13} = 20 \times 10^{12} \text{ K}^4

T^2 = \sqrt{20} \times 10^6 \approx 4.472 \times 10^6 \text{ K}^2

T = \sqrt{4.472} \times 10^3 \approx 2115 \text{ K}

Now, we calculate the exponent

\dfrac{\phi}{kT}

for the Richardson-Dushman equation:

\dfrac{\phi}{kT} = \dfrac{4.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23} \times 2115} \approx \dfrac{7.2 \times 10^{-19}}{2.918 \times 10^{-20}} \approx 24.67

The exponential factor is:

e^{-\phi / kT} = e^{-24.67} \approx 1.93 \times 10^{-11}

The saturation current

I

is given by the Richardson-Dushman equation multiplied by the surface area:

I = A S T^2 e^{-\phi / kT}

Substituting the values:

I = (60 \times 10^4) \times (2.0 \times 10^{-5}) \times (4.472 \times 10^6) \times (1.93 \times 10^{-11})

I = 12 \times 4.472 \times 10^6 \times 1.93 \times 10^{-11}

I \approx 53.66 \times 10^6 \times 1.93 \times 10^{-11} \approx 1.03 \times 10^{-3} \text{ A}

Rounding to the nearest given option, we get

I \approx 0.001 \text{ A}

.

Answer:

0.001 \text{ A}

For tungsten, the constant A in the Richardson–Dushman equation is...

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