Determine the ratio n(T)/n(1000 K) for a tungsten emitter at temperatures T = 300 K , 2000 K , and 3000 K . Here, n(T) denotes the number of thermions emitted per second by the surface at temperature T . The work

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 6.57 × 10^ -55 , 9.73 × 10^ 11 , 1.37 × 10^ 16

According to the Richardson-Dushman equation, the number of thermions emitted per second by a surface at temperature

T

is given by:

n(T) \propto T^2 e^{-\dfrac{W}{k_B T}}

The ratio of the number of thermions emitted at temperature

T

to that at

1000 \text{ K}

is:

\dfrac{n(T)}{n(1000)} = \left( \dfrac{T}{1000} \right)^2 e^{-\dfrac{W}{k_B} \left( \dfrac{1}{T} - \dfrac{1}{1000} \right)}

Given the work function

W = 4.52 \text{ eV}

and the Boltzmann constant

k_B \approx 8.62 \times 10^{-5} \text{ eV/K}

, we have:

\dfrac{W}{k_B} = \dfrac{4.52}{8.62 \times 10^{-5}} \approx 52436 \text{ K}

For

T = 300 \text{ K}

:

\dfrac{n(300)}{n(1000)} = \left( \dfrac{300}{1000} \right)^2 e^{-52436 \left( \dfrac{1}{300} - \dfrac{1}{1000} \right)}

\dfrac{n(300)}{n(1000)} = 0.09 \times e^{-122.35} \approx 0.09 \times 7.30 \times 10^{-54} = 6.57 \times 10^{-55}

For

T = 2000 \text{ K}

:

\dfrac{n(2000)}{n(1000)} = \left( \dfrac{2000}{1000} \right)^2 e^{-52436 \left( \dfrac{1}{2000} - \dfrac{1}{1000} \right)}

\dfrac{n(2000)}{n(1000)} = 4 \times e^{26.218} \approx 4 \times 2.43 \times 10^{11} = 9.73 \times 10^{11}

For

T = 3000 \text{ K}

:

\dfrac{n(3000)}{n(1000)} = \left( \dfrac{3000}{1000} \right)^2 e^{-52436 \left( \dfrac{1}{3000} - \dfrac{1}{1000} \right)}

\dfrac{n(3000)}{n(1000)} = 9 \times e^{34.957} \approx 9 \times 1.52 \times 10^{15} = 1.37 \times 10^{16}

Answer:

6.57 \times 10^{-55}

,

9.73 \times 10^{11}

,

1.37 \times 10^{16}

Determine the ratio n(T)/n(1000 K) for a tungsten emitter at...

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