Solving Double Square Well Potential - Physical Chemistry Homework

[ywkim880801]

this is one of my pchem class homework...
and I have no idea how to solve this...

compute $$\tau$$ for and electron in the double square well potential of width L = 2 Bohr, and depth $$\bar{V}$$ = 4 and separation R = 3. Use atomic units for your computation...

and in his lecture notes... he gave us (tau) = h/$$\DeltaE$$ .

he said it's just a simple plug-in problem... but i don't know where that $$\DeltaE$$

comes from...

please help me...

pchem sucks...

 

[azntoon]

The energy difference \DeltaE in this problem is the difference between the energies of the two quantum states that are separated by the double square well potential. To calculate this, you can use the equation for the energy of an electron in a one-dimensional box: E_n = n^2 \hbar^2 \pi^2 / (2mL^2)Where n is an integer indicating which quantum state the electron is in. For the lower state, n will be equal to 1, and for the higher state, n will be equal to 2. Thus, the energy difference \DeltaE between these two states is given by: \DeltaE = E_2 - E_1 = \hbar^2 \pi^2 / (2mL^2)Substituting in the given values of L, m, and \hbar, we get:\DeltaE = 4 \pi^2 \hbar^2 / (2 \cdot 2 \cdot (4 \pi \epsilon_0)^2 ) = 4 / (4 \pi \epsilon_0)^2Now that we have \DeltaE, we can plug it into the equation for \tau:\tau = h / \DeltaE = (4 \pi \epsilon_0)^2 / 4 = \pi^2 \epsilon_0^2