$q_{vib} (T) = \frac{e^{-h\nu/2k_BT}}{1-e^{-h\nu/k_BT}}$
For $\frac{h\nu}{k_BT} = 3$, the probability of finding the harmonic oscillator in its ground vibrational state is ____________ (Up to two decimal places)
Given
Canonical partition function of harmonic oscillator:
$q_{vib}(T) = \dfrac{e^{-h\nu/2k_BT}}{1 - e^{-h\nu/k_BT}}$
Given ratio: $\dfrac{h\nu}{k_BT} = 3$
Step 1: Write expression for ground state probability
The probability of finding the oscillator in the ground state is
$P_0 = \dfrac{e^{-h\nu/2k_BT}}{q_{vib}}$
Step 2: Substitute $q_{vib}$ into $P_0$
$P_0 = \dfrac{e^{-h\nu/2k_BT}}{\dfrac{e^{-h\nu/2k_BT}}{1 - e^{-h\nu/k_BT}}}$
This simplifies to
$P_0 = 1 - e^{-h\nu/k_BT}$
Step 3: Substitute numerical value
Given $\dfrac{h\nu}{k_BT} = 3$
$P_0 = 1 - e^{-3}$
$P_0 = 1 - 0.0498 = 0.9502$
Final Answer
$\boxed{0.95}$
Important Questions from Partition Functions and Their Relation
Six distinguishable particles are distributed over 3 non‐degenerate levels, of energies 0, ε and 2ε. The most probable value for the total energy is
The partition function for a gas is given by
Q(N, V, T) = \(\frac{1}{N!}\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}\) (v - Nb)Ne \(\frac{\beta aN^2}{V}\)
The internal energy of the gas is
A three-state system with energies E = −ε0, 0, +ε0 is in a thermal equilibrium at a temperature T. If β ε0 = x, the probability of finding the system with energy E = 0 is [recall, cosh x = \(\frac{1}{2}\)(ex + e−x)]
The translational, vibrational, and rotational molecular partition functions for a system containing ideal diatomic gas molecules in the canonical ensemble (N, V, T) are written as, $q_{trans}$, $q_{vib}$, and $q_{rot}$, respectively. The option that correctly defines their thermodynamic variable(s) dependency is
If $q_t$ and $Q_{t,m}$ are the molecular and molar translational partition functions of $X_2$, respectively, then $ln(Q_{t,m})$ =
(N is the Avogadro number)
