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Posted

hey everyone,

 

I am doing QM, trying to solve as much problems there are to get a strong hold of the idea. here one question which i cant understand, i think that there is not enough information to solve this one;

 

Assuming that a chemical bond can be modelled as a one-dimensional box,estimate the order of magnitude of the de Broglie wavelength of an electron in a typical chemical bond.

 

do you think we have enough information in the question to solve this. The de broglie wavelength can be calculated using [math] lambda = \frac {h}{p} [/math] where p is momentum. how do you calculate momentum? I dont get it. Could anyone help. thanks!.

Posted

It does say to estimate it, rather than calculate it. try this: work with a hydrogen-hydrogen bond. what is the length of it? how many nodes are there between the nuclei? given the number of nodes, how big is the wavelength? are most bonds longer or shorter? by how much?

 

I may be going down the wrong route but that's how I'd approach it.

Posted (edited)

what I think is that if we find the energy of electron in 1D box, by assuming that the length of box is [math] 10^{-10} m [/math] and n=1 (lowest energy of the system) then we can find the momentum but we we would have to use the classical foruma relating kinetic energy with momentum;

 

[math] E= \frac {p^2}{2m} [/math]

 

if the use of the above forumla is acceptable ( i say acceptable because then it will be switching from QM to classical mechanics), then we can find mommentum and then use the de broglie relation to get the wavelegnth?

 

 

BTW just did the calculation assuming what i said above is correct, and the de broglie comes out [math] 1.99*10^{-10} m[/math]

Edited by ChemSiddiqui
Posted
what I think is that if we find the energy of electron in 1D box, by assuming that the length of box is [math] 10^{-10} m [/math] and n=1 (lowest energy of the system) then we can find the momentum but we we would have to use the classical foruma relating kinetic energy with momentum;

 

[math] E= \frac {p^2}{2m} [/math]

 

if the use of the above forumla is acceptable ( i say acceptable because then it will be switching from QM to classical mechanics), then we can find mommentum and then use the de broglie relation to get the wavelegnth?

 

 

BTW just did the calculation assuming what i said above is correct, and the de broglie comes out [math] 1.99*10^{-10} m[/math]

 

And if your particle is in a box of length L, the lowest state is a half-wave fitting into it, so the wavelength will be 2L (or a bit longer, depending on the depth of the well, since the tail of the wave goes into the sides a little), which is exactly what you've shown, and what hermanntrude suggested.

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