Transitional state of electrons.

  • Thread starter airkapp
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This is how I went about it but I am unsure if it is correct..
An n=3 to n=1 transition from an electron trapped in a rigid box produces a 240 nm photon. What is the width of the box?


λ = hc / E:

E = hc / λ

E = (1.06E-34 J*s * 3E8m/s) / 240nm

= 1.33E-19 J

E = h2 / 8mL2

L2 = h2 / 8mE
L = √ h2 / 8mE

L = √ ((1.06E-34 J*s) 2 / (8*9.11E-31kg*1.33E-19 J))

= 1.08E-10 m
 

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  • #2
Andrew Mason
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airkapp said:
This is how I went about it but I am unsure if it is correct..
An n=3 to n=1 transition from an electron trapped in a rigid box produces a 240 nm photon. What is the width of the box?
You are missing a factor of 8:

Allowed values of electron momentum are:

[tex]p = \frac{hn}{2L}[/tex]

Since [itex]E = p^2/2m[/itex], the allowed energy values are:

[tex]E = \frac{n^2h^2}{4L^2}\frac{1}{2m} =n^2\frac{h^2}{8mL^2}[/tex]

The change in energy between n=3 and n=1 results in the photon (E=hv):

[tex]\Delta E = (9-1)\frac{h^2}{8mL^2} = \frac{hc}{\lambda}[/tex]

[tex]L^2 = 8\frac{h\lambda}{8cm}[/tex]

[tex]L = \sqrt{\frac{h\lambda}{cm}}[/tex]

AM
 

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