Heisenberg and his energy

[Sarahisme]

Hey all,

i can do part (a) (or at least i think i can, i get: 9.4eV

 

but for part (b), i can't find a formula to use. any ideas guys/gals?

 

and for part © i caluclated that you would need to give ~28eV worth of energy to the an elecron to excite it to the first excited state. i am unsure whether this is said to be 'easy'.... do you think 'easy' involves cmplicated equipment or me poking it with my finger?

 

 

sorry bout the 1000 questions

 

Sarah

[Sarahisme]

hmm ok maybe i have something here...

 

can you just use the formula:

 

[math] KE \geq \frac{\hbar^2}{8(\Delta x)^2 m} [/math]

 

as for part ©

 

i guess that heating up say a sodium salt which glows yellow when put in a bunsen flame is an example of 'easy' excitment of an electron to the next excited state.?

[Severian]

OK, lets take a look.

 

a) you have done (I presume correctly - I am not going to put in the numbers).

 

b) The Heisenburg Uncertainty Principle is telling us that if we know the position of the particle very well then its energy is rather uncertain (and vice versa). Do we know the position of the particle in this example very well? Can you make any statement at all about where it is? And if you can, what does this say about its energy?

 

c) Notice the statement about room temperature? What would the kinetic energy of the particle be if it was at room temperature? How does this compare with the energy difference between the two levels? And what does this mean?

[Sarahisme]

b) The Heisenburg Uncertainty Principle is telling us that if we know the position of the particle very well then its energy is rather uncertain (and vice versa). Do we know the position of the particle in this example very well? Can you make any statement at all about where it is? And if you can' date=' what does this say about its energy?[/quote']

 

yep i understand.

 

c) Notice the statement about room temperature? What would the kinetic energy of the particle be if it was at room temperature? How does this compare with the energy difference between the two levels? And what does this mean?

 

c) Thermodynamics question' date=' think E=kT.[/quote']

 

oh i see, so i should use that average KE = (1/2)kT and not 3/2, because we are only using the one-dimensional infinite square well model to find the energies of the two different levels?

 

k= boltzmann's constant

 

so you can work out the temperature increase needed to eaise the energy by enough to push the electron into the first excited state?

 

i.e. T = 2(E2-E1)/k

?

 

lol, but then for some reason i get that i would need an increase in temperature of ~ 650087 K ....think i may have stuffed things up somewhere

 

this seems a bit much considering you only need to put in ~4.5 x 10^{-18} J of energy to raise it to the first excited state...

[timo]

I had approached the 3rd question slightly differently but it´s about a rough approx so there´s no "the one" way to do it. The 650 kK seem pretty reasonable to me:

In semiconductors, the energy gap between valence band and conducting band is in the order of eV (10^-19 J), IIRC. At room temperature, only a tiny fraction of the electrons are in the conducting band, so thermal energies at 300 K are much below that energy gap.

[Sarahisme]

yep ok i see. so its the right way to do the calulation, but i can see that it is quite a rough approximation.

 

Thank you for your help Atheist

 

-sarah