Question

The mean energy of a molecule having two available energy states at \( \epsilon = 0 \) J and \( \epsilon = 4.14 \times 10^{-21} \) J at 300 K is ___________ \( \times 10^{-21} \) J (rounded off to two decimal places). [Given: Boltzmann constant \( k_B = 1.38 \times 10^{-23} \) J K\(^{-1}\)]

Hint:

The mean energy is a weighted average of the available energy states, where the weights are given by the Boltzmann factors \( e^{-\epsilon_i / (k_B T)} \), which represent the probability of a molecule being in a particular energy state. At higher temperatures, higher energy states become more populated, and the mean energy increases.

Answer 1

The mean energy of a molecule in a system with discrete energy levels is given by the Boltzmann distribution: \[ \langle \epsilon \rangle = \frac{\sum_i \epsilon_i e^{-\epsilon_i / (k_B T)}}{\sum_i e^{-\epsilon_i / (k_B T)}} \] where \( \epsilon_i \) are the energy levels, \( T \) is the temperature, and \( k_B \) is the Boltzmann constant. In this case, there are two energy states: \( \epsilon_1 = 0 \) J \( \epsilon_2 = 4.14 \times 10^{-21} \) J The temperature is \( T = 300 \) K. The Boltzmann constant is \( k_B = 1.38 \times 10^{-23} \) J K\(^{-1}\). First, calculate \( k_B T \): \[ k_B T = (1.38 \times 10^{-23} \text{ J K}^{-1}) \times (300 \text{ K}) = 4.14 \times 10^{-21} \text{ J} \] Now, substitute the values into the mean energy formula: \[ \langle \epsilon \rangle = \frac{(0 \cdot e^{-0 / (4.14 \times 10^{-21})}) + (4.14 \times 10^{-21} \cdot e^{-(4.14 \times 10^{-21}) / (4.14 \times 10^{-21})})}{e^{-0 / (4.14 \times 10^{-21})} + e^{-(4.14 \times 10^{-21}) / (4.14 \times 10^{-21})}} \] \[ \langle \epsilon \rangle = \frac{(0 \cdot e^0) + (4.14 \times 10^{-21} \cdot e^{-1})}{e^0 + e^{-1}} \] \[ \langle \epsilon \rangle = \frac{0 + (4.14 \times 10^{-21} \cdot \frac{1}{e})}{1 + \frac{1}{e}} \] The value of \( e \approx 2.71828 \). \[ \frac{1}{e} \approx \frac{1}{2.71828} \approx 0.36788 \] \[ \langle \epsilon \rangle = \frac{4.14 \times 10^{-21} \times 0.36788}{1 + 0.36788} = \frac{1.5220392 \times 10^{-21}}{1.36788} \] \[ \langle \epsilon \rangle = 1.1127 \times 10^{-21} \text{ J} \] The mean energy is \( 1.1127 \times 10^{-21} \) J. The question asks for the value in the form \underline{\hspace{2cm}} \( \times 10^{-21} \) J, rounded off to two decimal places. So, the value is 1.11. This value (1.11) falls within the given correct answer range of 1.00 to 1.20.