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Problem II A vibronic transition consist of concerted electronic and vibrational excitations. First few vibronic excitations from zeroth to respectively θ = 2,3,4,5 states are given by v= 12569.95...
4. Anharmonic potential (15 points) The adjacent figure shows the experimentally determined potential energy curve of the electronic ground state of"Br2, with a few of the vibrational levels. The...
4. Anharmonic potential (15 points) The adjacent figure shows the experimentally determined potential energy curve of the electronic ground state of"Br2, with a few of the vibrational levels. The vibrational transitions are reasonably well described by a harmonic oscillator model, but much more accurately by including a small anharmonic correction term: En/hcVe(n 1/2) - vexe(n + 1/2)2. From fits to experimental data, the values of the constants are 325.32 cm and exe 1.08 cm .5 10 15 (a) Calculate the...
Problem 4.1 - Odd Bound States for the Finite Square Well Consider the finite square well...
Problem 4.1 - Odd Bound States for the Finite Square Well Consider the finite square well potential of depth Vo, V(x) = -{ S-V., –a sx sa 10, else In lecture we explored the even bound state solutions for this potential. In this problem you will explore the odd bound state solutions. Consider an energy E < 0 and define the (real, positive) quantities k and k as 2m E K= 2m(E + V) h2 h2 In lecture we wrote...
4. Anharmonic potential (15 points) The adjacent figure shows the experimentally determined potential energy curve of the electronic ground state of"Br2, with a few of the vibrational levels. The vibrational transitions are reasonably well described by a harmonic oscillator model, but much more accurately by including a small anharmonic correction term: En/hcVe(n 1/2) - vexe(n + 1/2)2. From fits to experimental data, the values of the constants are 325.32 cm and exe 1.08 cm .5 10 15 (a) Calculate the...
Problem 4.1 - Odd Bound States for the Finite Square Well Consider the finite square well potential of depth Vo, V(x) = -{ S-V., –a sx sa 10, else In lecture we explored the even bound state solutions for this potential. In this problem you will explore the odd bound state solutions. Consider an energy E < 0 and define the (real, positive) quantities k and k as 2m E K= 2m(E + V) h2 h2 In lecture we wrote...
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