The vibrational frequency of a homonuclear diatomic molecule is v. The...
Solution:
The energy of a homonuclear diatomic molecule is given by:
E(v) = (v + 1/2) hν
Where, v is the vibrational quantum number and h is the Planck's constant.
At a given temperature T, the population of a given vibrational level i is given by:
Ni/N0 = (gi/g0) exp(-Ei/kBT)
Where, gi and g0 are the degeneracies of the excited and ground states respectively, Ei is the energy of the excited state, and kB is the Boltzmann constant.
To find the temperature at which the population of the first excited state will be half that of the ground state, we need to solve the following equation:
N1/N0 = 1/2
(g1/g0) exp(-E1/kBT) = 1/2
Taking the natural logarithm of both sides, we get:
ln(g1/g0) - E1/kBT = ln(1/2)
ln(g1/g0) + E1/kB(T1/2) = ln(g1/g0) + E1/kBT
E1/kB(T1/2) = ln(2)
T1/2 = (ln2/kB) (E1/h)
Substituting the energy of the first excited state, E1 = 2hν, we get:
T1/2 = hv/(ln2.kB)
Therefore, the correct answer is option B.
The vibrational frequency of a homonuclear diatomic molecule is v. The...
According to Boltzmann's population formula
N/n=exp(-E/kT)
As per question N=n/2 so
0.5=exp(-E/kT)
taking ln both side
ln(0.5)= -E/kT
ln(2)=hv/kT
so T=hv/(ln2.k)
where N=population in excited state
n=population in ground state