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Answer :
To solve this problem we will use the concept related to electrons in a box which determines the energy of an electron in that state.
Mathematically this expression is given by,
[tex]E_n= \frac{n^2h^2}{8mL^2}[/tex]
Where,
m = mass of an electron
h = Planck's constant
n = is the integer number of the eigenstate
L = Quantum well width
The change in energy must be given in state 1 and 2, therefore
[tex]\Delta E = E_2 - E_1[/tex]
[tex]\Delta E = \frac{2^2h^2}{8mL^2}-\frac{1^2h^2}{8mL^2}[/tex]
[tex]\Delta E = \frac{3h^2}{8mL^2}[/tex]
Replacing we have:
[tex](4*1.6*10^{-19}) = \frac{3(6.626*10^{34})}{8*(9.11*10^{-31})*L^2}[/tex]
[tex]L = 0.53nm[/tex]
Therefore the correct answer is C.
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