The question asks us to find the thermal voltage at a specific temperature. The thermal voltage ($V_T$) is a concept often used in semiconductor physics and electronics. It represents the voltage equivalent of thermal energy ($kT$) per charge carrier ($q$).
The formula to calculate thermal voltage is:
$ V_T = \frac{k_B T}{q} $Where:
The temperature is given in Celsius (°C), but the formula requires the temperature in Kelvin (K). The conversion formula is:
$ T(\text{K}) = T(\text{°C}) + 273.15 $Given the temperature is 47°C:
$ T = 47 + 273.15 = 320.15 \text{ K} $Now, substitute the values into the thermal voltage formula:
$ V_T = \frac{(1.38 \times 10^{-23} \text{ J/K}) \times (320.15 \text{ K})}{1.602 \times 10^{-19} \text{ C}} $First, calculate the numerator:
$ k_B T = (1.38 \times 10^{-23}) \times 320.15 \approx 4.41807 \times 10^{-21} \text{ J} $Now, divide by the elementary charge ($q$):
$ V_T = \frac{4.41807 \times 10^{-21} \text{ J}}{1.602 \times 10^{-19} \text{ C}} $Performing the division:
$ V_T \approx 0.02758 \text{ V} $The options are given in millivolts (mV). To convert Volts (V) to millivolts (mV), multiply by 1000:
$ V_T (\text{mV}) = 0.02758 \text{ V} \times 1000 \text{ mV/V} $ $ V_T \approx 27.58 \text{ mV} $Rounding the result to one decimal place gives 27.6 mV.
Let's compare our calculated result with the given options:
Our calculated value of approximately 27.6 mV matches Option 1.