Transitional state of electrons.

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An electron transition from n=3 to n=1 in a rigid box emits a 240 nm photon, leading to a calculated energy of 1.33E-19 J. The width of the box is derived using the formula L = √(h² / (8mE)), resulting in an approximate width of 1.08E-10 m, or 108 picometers. Corrections were noted regarding the inclusion of factors in the momentum and energy equations. The calculation assumes a one-dimensional box, and variations in dimensions would alter the result. The discussion emphasizes the importance of accurate formulas in quantum mechanics for determining physical properties.
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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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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:

p = \frac{hn}{2L}

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

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

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

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

L^2 = 8\frac{h\lambda}{8cm}

L = \sqrt{\frac{h\lambda}{cm}}

AM
 


Therefore, the width of the box is approximately 108 picometers. This calculation assumes a one-dimensional box, as the question does not specify the dimensions of the box. If the box is two or three-dimensional, the calculation would change accordingly.
 
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