(6-12)  ①  We know that (classically) Ek  =  ½ mv2   =  ½ x 2.5 x 2.02  =  5.0 J

②  Rearranging, we get:

=  1.5 x 1035

So he’s certainly not going to be in his ground state:  he’s going to be in a very high quantum level.

③  Cormorant’s momentum will be: p = mv  =  2.5 x 2.0  =  5.0 kgms-1

The de Broglie equation then gives  λ = h/p   = 6.63x10-34 / 5  =   1.3 x 10-34 m

Now, looking at the particle in a box wavefunctions:

You can see that the number of full wavelengths that fit into the box is n/2. 

Now look at what happens as we increase the value of the quantum number, n, more drastically:

If we consider the probability of the particle being located in a short segment of length Δ, you can see that when the quantum number is small, the answer is very dependent on where the segment is placed:  in the n=1 state, the probability (which, remember, is proportional to the square of the wavefunction value) is much larger at position a than at b, whereas the reverse is clearly true at n=10.

Once we get to higher quantum numbers, however, where Δ is comparable to the wavelength, the probability will be similar everywhere. In the n=200 case, above, you can see that it will be virtually constant anywhere in the box. 

In Cormorant’s case, we've estimated the de Broglie wavelength as 1.3 x 10-34 m. So there will be 10/1.3 x 10-34  =  7.7x1034 wavelengths along the length of the box. That number is just huge, so the effect of the oscillations in the wave function will be way beyond negligible:  we can say it’s the same everywhere in the box, exactly as classical physics would predict.

④  The energy gap between adjacent energy levels is given by:

(n+1)2 – n2  =  2n+1

Since 2n>>1, in this case, we can confidently approximate this as 2n  (= 3.0 x 1035)

So we get

=  6.6 x 10-35 J

This energy gap is so tiny (remember that Cormorant’s initial kinetic energy was 5 J) as to be entirely negligible. In other words there is, to all intents and purposes, a continuum of possible energy levels accessible to Cormorant. Again, the quantisation is just not going to be detectable.

⑤   At around 300 K (which is close to room temperature), ½ kBT = 2.1 x 10-21 J.

This is nearly 500 times smaller than the energy needed to promote an electron to n=2 which means there is almost no chance of the electron making it. So we can confidently predict that the electrons in buta-1,3-diene will almost all be sitting in their ground states, with the slim chance of a UV photon coming by, their only hope of a bit of excitement.