Question

The force constant for HF is 966 N m^-1. Using the harmonic oscillator model, calculate the relative population of the first excited state and the ground state at 300 K.

Answer 1

To solve this problem it is necessary to apply the concepts related to the force constant of HF with energy excitation. Said expression may be mathematically encompassed as follows.

$hv = \frac{h}{2\pi}\sqrt{\frac{k}{\mu}}$

Here

h = Planck's Constant

v = Frequency of radiation required for excitation

k = Force constant

$\mu$ = Reduced mass of the molecule

The relative population of two energy levels is expressed by the relation as follows

$\frac{P(n_1)}{P(n_2)} = e^{\frac{hv}{K_bT}}$

Here

$P_{(n_1),(n_2)}$ = Population of energy at each level

$K_b$ = Boltzmann constant

T= Temperature of the molecule (300K at our case)

Calculate the reduced mass of HF molecule as follows,

$\mu = \frac{m_Hm_F}{m_H+m_F}$

$\mu = \frac{1amu\times19amu}{(1+19)amu}$

$\mu = \frac{19}{20}amu$

Convert the reduced mass in terms of kg,

$\mu = \frac{19}{20}amu (\frac{1g}{6.022*10^{23}amu})(\frac{1kg}{1000g})$

$\mu = 1.58*10^{-27}kg$

Now the excitation energy for two energy states would be

$hv = \frac{6.626*10^{-34}J\cdot s^{-1}}{2*3.14}\sqrt{\frac{966Nm^{-1}}{1.58*10^{-27}kg}}$

$hv = 1.05*10^{-34}J\cdot s^{-1} \sqrt{\frac{966kg \cdot s^2}{1.58*10^{-27}}kg}$

$hv = 8.24*10^{-20}J$

Then calculating the relative population of two energy states we have that

$\frac{P(n_1)}{P(n_2)} = \frac{P(excited)}{P(ground)} = e^{\frac{-8.24*10^{-20}J}{1.380*10^{-23}J/K*300K}}$

$\frac{P(n_1)}{P(n_2)} = e^{\frac{-8.24*10^{-20}}{4.14*10^{-21}}}$

$\frac{P(n_1)}{P(n_2)} = e^{-19.9}$

$\frac{P(n_1)}{P(n_2)} = 2.28*10^{-9}$

Therefore the relative population of excited stated to ground energy state is $2.28.10*10^{-9}$