One crude method of determining the size of a molecule is to treat the molecule as an infinite square well with an electron trapped inside, and to measure the wavelengths of emitted photons. If the photon emitted during the n = 2 to n =1 transition has wavelength 1940.0 nm, what is the width of the molecule? Is it simply `sqrt((3hlambda)/(8mc)) = 1.3nm`

Expert Answers

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The formula seems correct. It can be obtained from the following facts:

`E_n = (n^2 h^2)/(8mL^2)`  and  `lambda = (hc)/(E_2-E_1),`

where `n` is the state number (changes from `2` to `1` in our case),
`h` is the Planck's constant,
`m` is the mass of electron,
`L` is the size of a molecule,
`lambda` is the photon's wavelength.

Therefore `E_2-E_1 =((2^2-1^2) h^2)/(8mL^2) = (3 h^2)/(8mL^2),` and

`L^2 = (3h^2)/(8m(E_2-E_1)) = (3h^2)/(8m(hc)/lambda) = (3h lambda)/(8mc).`

Thus `L = sqrt((3h lambda)/(8mc)).` To find the numerical result, recall the values in standard units:

`h = 6.6*10^(-34),`  `m = 9.1*10^(-31),`  `c = 3*10^8,`  `lambda = 1.94*10^(-6).`

So the result is

`sqrt((3*6.6*1.94)/(8*9.1*3)*(10^(-34)*10^(-6))/(10^(-31)*10^8)) approx sqrt(0.176*10^(-17)) = sqrt(1.76*10^(-18)) approx 1.33*10^(-9).`

This value is in standard units, meters. Because nano means `10^(-9),` this is the same as `1.33 nm.` So your numerical answer is also correct:)

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