Question

Consider the quantum mechanical vibration of H2 in the n = 1 state. Calculate the expectation value of the potential energy, (Epotential), of this vibrational state. The wavefunction for the n = 1 state is: a = 6 where a = kr for Hz is 575 N/m and u = 8.368 x 10-20 kg The potential energy operator for the quantum harmonic oscillator is: Epotential = ***

Answer 1

The complete step wise calculation of the average potential energy is given as below:-

It is given that, wave function for n=1 state is givenby: - where, a= ( 3 hari ya kg = 575 Nim arkf= spring constant u= 8.368 810 28, K = Reduced mass and Epotential = kx² ,x=position of particle. t = Reduced planck's Constant=1.06810 3J. and I = 3.14.

Expectation value of potential energy (E) can be defined as below:- too LẼ - * () Ý, 44 – ® co whore, 4* is a complex Conjugate of 4, Then LE) - Sto Contrera) *(3) zett ) x leurrea) C D xé e de

on <ê) - Curiera) ** £)SI 34 de since, to a paxe dx = 1 For from n= even equation ③ we have Then

in 12 23 since [12 = da - and 3 = 3 x since (5/21-2, Then k XNT 3K x1223 423 on (e) = 3* x[51]* = 34(««2)¥}? on LÊ> = 34 x 25 = 34 22 putting the value of a and, we get.

Ê > = 3a V2 = 3 (tr) Ve _O putting all the values in the above equation, <Ê ) = 3x(106x1334) (57 458 21328) Y J or LÊS = 7.95 xl-35 (68.71) 12 x1014 J onê) = 795 x 8.29 x16-21 5 = 65.91x16-215 on LE) = 6.591 X10-20 J.