Question

At room temperature (293 K), calculate ${\mathbf{k}}_{\mathbf{B}}\mathbf{T}$ in joules and eV.

Short answer

The value of $k_B\mathrm{T}$is $4.04\times {10}^{-21}\mathrm{J}$and the value of$k_{B}T$eV is role="math" localid="1657860404123" $2.52\times {10}^{-2}\mathrm{eV}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The value of room temperature is,${\mathrm{T}}_{\mathrm{R}}=293\mathrm{K}$.
• The value of Boltzmann constant is,$k_B=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

Concept/Significance of kBT.

${\mathbf{k}}_{\mathbf{B}}\mathbf{T}$has units of energy and hence in general it’s just a metric to quantify energy Instead of utilizing units of joules, calories etc. which are more practical at a macroscopic level${\mathbf{k}}_{\mathbf{B}}\mathbf{T}$is preferred unit

Determination of the value of the product of Boltzmann constant and temperature in joules and eV

The value of the value of the product of Boltzmann constant and temperature in joule is given by,

$$\begin{gathered} k_{\mathrm{B}}\mathrm{T}=(293K)(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K} \\ =4.04\times {10}^{-21}\mathrm{J} \end{gathered}$$

Thus, the value of $k_{\mathrm{B}}\mathrm{T}$is$4.04\times {10}^{-21}\mathrm{J}$.

The value of the product of Boltzmann constant and temperature in eV is given by,

$$\begin{gathered} k_{\mathrm{B}}\mathrm{T}=4.04\times {10}^{-21}J\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right) \\ =2.52\times {10}^{-2}\mathrm{eV} \end{gathered}$$

Thus, the value of$k_{\mathrm{B}}\mathrm{T}$eV is $2.52\times {10}^{-2}\mathrm{eV}$.