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 = (h)/(2\pi)\sqrt{(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

`(P(n_1))/(P(n_2)) = e^{(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 = (m_Hm_F)/(m_H+m_F)`

`\mu = (1amu*19amu)/((1+19)amu)`

`\mu = (19)/(20)amu`

Convert the reduced mass in terms of kg,

`\mu = (19)/(20)amu ((1g)/(6.022*10^(23)amu))((1kg)/(1000g))`

`\mu = 1.58*10^(-27)kg`

Now the excitation energy for two energy states would be

`hv = (6.626*10^(-34)J\cdot s^(-1))/(2*3.14)\sqrt{(966Nm^(-1))/(1.58*10^(-27)kg)}`

`hv = 1.05*10^(-34)J\cdot s^(-1) \sqrt{(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

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

`(P(n_1))/(P(n_2)) = e^{(-8.24*10^(-20))/(4.14*10^(-21))}`

`(P(n_1))/(P(n_2)) = e^(-19.9)`

`(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)`