5.
A protein can exist in two conformational states A and B with energies -E and +E
respectively. Assume the system (single protein) is in thermal equilibrium with a heat
bath at temperature T. The average energy in the infinite temperature case and zero
temperature case are respectively given by
a. E, -E
b. ∞, 0
c. E, -2E
d. 0, -E
A protein exists in two states, A (-E energy) and B (+E energy), in thermal equilibrium. The average energy at infinite temperature is 0, while at zero temperature it is -E. The correct option is d. 0, -E
Boltzmann Distribution Basics
The partition function \( Z = e^{\beta E} + e^{-\beta E} = 2 \cosh(\beta E) \), where \( \beta = 1/(kT) \). Average energy \( \langle E \rangle = \frac{-E e^{\beta E} + E e^{-\beta E}}{Z} = -E \tanh(\beta E) \).
Infinite Temperature Limit
At infinite temperature (\( T \to \infty \), \( \beta \to 0 \)), states A and B become equally probable (\( p_A = p_B = 1/2 \)). Thus, \( \langle E \rangle = \frac{1}{2}(-E) + \frac{1}{2}(+E) = 0 \).
Zero Temperature Limit
At zero temperature (\( T \to 0 \), \( \beta \to \infty \)), the system occupies only the ground state A (\( p_A = 1 \), \( p_B = 0 \)). Thus, \( \langle E \rangle = -E \).
Option Analysis
- a. E, -E: Wrong; infinite T yields 0, not E.
- b. ∞, 0: Wrong; energies are finite, zero T gives lowest energy (-E).
- c. E, -2E: Wrong; no state has -2E, infinite T is 0.
- d. 0, -E: Correct, matches limits derived from statistical mechanics.
