Determine the factor by which the emission current changes when the...

According to the Richardson-Dushman equation, the emission current $I$ is given by:

$I = A T^2 e^{-\dfrac{W}{kT}}$

Taking the natural logarithm on both sides:

$\ln I = \ln A + 2 \ln T - \dfrac{W}{kT}$

Differentiating with respect to $T$:

$\dfrac{dI}{I} = \dfrac{2}{T} dT + \dfrac{W}{kT^2} dT$

Given values are $T = 2000 \text{ K}$, $dT = 10 \text{ K}$, $W = 4.5 \text{ eV}$, and Boltzmann constant $k \approx \dfrac{1}{11600} \text{ eV/K}$. Substituting these values:

$\dfrac{dI}{I} = \dfrac{2 \times 10}{2000} + \dfrac{4.5 \times 11600 \times 10}{(2000)^2}$

$\dfrac{dI}{I} = 0.01 + \dfrac{522000}{4000000}$

$\dfrac{dI}{I} = 0.01 + 0.1305 = 0.1405$

The new emission current is $I' = I + dI = I(1 + 0.1405) = 1.1405 I$.

Thus, the emission current changes by a factor of approximately $1.14$.