A Hint of Quantum Theory — Question 7

Question 7

Changing isotope mass shifts the temperature scale for vibration

Two isotopic molecular models have the same bond stiffness but reduced masses $\mu_A$ and $\mu_B=4\mu_A$. The mode frequency is $\omega=\sqrt{\kappa/\mu}$ and $\Theta=\hbar\omega/k_B$. For a quantum harmonic mode use $C/R=x^2e^x/(e^x-1)^2$, $x=\Theta/T$.

1. Find $\Theta_B/\Theta_A$ and compare both vibrational capacities at the common temperature $T=\Theta_A/2$.

2. If a chosen capacity fraction is reached by A at temperature $T_A$, find the temperature at which B reaches the same fraction. Explain why the classical high-temperature capacities nevertheless agree.

Question 7 - Solution

Changing isotope mass shifts the temperature scale for vibration

(a) Method and calculation

$$\Theta_B/\Theta_A=\omega_B/\omega_A=\sqrt{\mu_A/\mu_B}=\tfrac12.$$$$T=\Theta_A/2:\quad x_A=2,\ x_B=1,\quad C_A/R\approx.724062,\ C_B/R\approx.920674.$$

(b) Interpretation and further calculation

$$C_B(T_B)=C_A(T_A)\quad\text{at}\quad\frac{\Theta_B}{T_B}=\frac{\Theta_A}{T_A}\ \Rightarrow\ T_B=T_A/2.$$$$x\to0:\quad\frac{x^2e^x}{(e^x-1)^2}\to1\ \Rightarrow\ C_A,C_B\to R.$$A heavier reduced mass reduces the spacing, enabling excitation at lower temperature without changing the classical number of quadratic terms.

Physical check

The reduced mass controls bond vibration; it is not generally the total molecular mass.