The square of the vibrational wavefunction for a diatomic molecule gives the pro

Question

The square of the vibrational wavefunction for a diatomic molecule gives the probability P for finding the atoms separated by a distance R + x, where R is the equilibrium distance. For H^35 Cl, assume R = 1.27 A and the force constant is 4.82 times 10^2 N m^-1. What are the 2 most probable H-Cl distances for H^35 Cl in its 1st excited vibrational state (v = 1)? The wavefunction is given below (Atkins 8.27), where H_v is a Hermite polynomial, is the reduced mass, k_f is the force constant, and N is a normalization coefficient. psi_v(x) = N_v H_v(y)e^-y^2/2 where y = x/alpha and alpha = (h^2/mu k_f)^1/4

Explanation / Answer

First of all vibrational wavefunction

(x)=NvHv(y)exp(-y^2/2)

now y=x/alpha

alpha =(hcut/meu*Kf)^1/4 meu=.979amu for hcl

usin values alpha= (1.05*10^-27cm^2g/s/[.979*1.6*10^-27*4.82*10^2] )^1/4= .0000868

then y=x/alpha=x/ .0000868

substituting this in the wavefunction expression we get

(x)=NvHv(x/ .0000868)exp(-x/ .0000868^2/2)

(x)=NvHv(x/ .0000868)exp(-x^2/.000000015)

now probability of finding atoms seperated by R+x =1.27+x is P

and according to the question

P=(x)^2 =Nv^2Hv^2(x/ .0000868)^2exp2(-x^2/.000000015)

Now at ground state the probability function shows the atoms position in maximum point

that means for 1st excited state or l=1 state we differentiate the probability function

therefore d(x)^2/dx= 2(x)= NvHv(- .0000868x-1)exp(-2x/.000000015)/ .0000868

putting x=1.27 we get

probable distance=(-1.00011) NvHvexp(-2.54)/ .0000868

again at minimum putting x=0 we get probable distance =NvHv

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