Question

Show that for a room-temperature semiconductor with a band gap of $1eV$, a temperature rise of 4K would raise the conductivity by about 30%.

Short answer

The rise in conductivity of the semiconductor is$30\%$ .

Step-by-step solution

Given information

The band gap of the semiconductor is, $E_g=1\text{eV}$.

The initial room temperature is, $T_i=300\text{K}$.

The temperature rise is$T_f-T_i=4\text{K}$,.

The final temperature is, $T_f=304\text{K}$.

Conductivity of semiconductors

For a semiconductor, the value of its conductivity relies upon the room temperature as well as the band gap energy of the material.

The formula showing the relationship between the conductivity and the temperature of the room is given below,

$N_{\text{excited}}=D{k}_{B}T{e}^{-\left(\frac{E_g}{2k_{B}T}\right)}$

Here, D is the constant, T is the temperature of semiconductor, $E_g$is the energy band gap of the semiconductor, and$k_B$ is the Boltzmann onstant.

Increase in conductivity

The ratio of the final conductivity to the initial conductivity of the semiconductor in terms of the ratio between the initial and final temperature can be calculated by using following relation,

$\frac{N_f}{N_i}=\frac{T_f{e}^{-\left(\frac{E_g}{2k_B{T}_f}\right)}}{T_i{e}^{-\left(\frac{E_g}{2k_B{T}_i}\right)}}$

Here,$N_f$is the final number of electrons in conduction band,$N_i$is the initialof electrons in conduction band, and $k_B$is the Boltzmann constant, its value is$1.38\times {10}^{-23}\text{J/K}$.

Multiplying $\frac{T_i}{T_f}$both sides,

$\begin{array}{rcl}\frac{T_i}{T_f}\times \frac{N_f}{N_i}& =& \frac{T_i}{T_f}\times \frac{T_f}{T_i}{e}^{-\left(\frac{E_g}{2k_B{T}_f}\right)+\left(\frac{E_g}{2k_B{T}_i}\right)}\\ \frac{T_i}{T_f}\times \frac{N_f}{N_i}& =& e^{\frac{E_g\left(T_f-T_i\right)}{2k_B{T}_i{T}_f}}\\ & & \end{array}$

Putting the values,

$$\begin{gathered} \frac{300\text{K}}{304\text{K}}\times \frac{N_f}{N_i}=e^{\frac{1.6\times {10}^{-19}\text{J}}{2\times 1.38\times {10}^{-23}\text{J/K}}\times \frac{4\text{K}}{300\text{K}\times 304\text{K}}} \\ 0.987\times \frac{N_f}{N_i}=e^{2.54\times {10}^{-5}\times {10}^4} \\ 0.987\times \frac{N_f}{N_i}=e^{0.254} \end{gathered}$$

Solving further,

$$\begin{gathered} \frac{N_f}{N_i}=\frac{1.29}{0.987} \\ {N}_f=1.3N_i \end{gathered}$$

So, the rise in conductivity of the semiconductor is given by,

$$\begin{gathered} \frac{N_f-N_i}{N_i}\times 100=0.254 \\ \frac{N_f-N_i}{N_i}\times 100=\frac{1.3N_i-N_i}{N_i}\times 100 \\ \frac{N_f-N_i}{N_i}\times 100=0.3\times 100 \\ \frac{N_f-N_i}{N_i}\times 100=30\% \end{gathered}$$

Hence proved, the rise in conductivity is$30\%$