The width of the box in order for the n = 2 to n = 1 is : 3.3 * 10⁻¹⁰ m
Given data :
λ = 122 nm
Transition ; n = 2 to n = 1
Determine the width of the box applying the formulae below
Total energy ( E₁ ) = [tex]\frac{h^{2} }{8mL^{2} }[/tex] --- ( 1 )
Change in energy = ΔE = [tex]\frac{hc}{λ}[/tex]
Next step : Determine the difference in Total energy and change in energy
ΔE - E₁ = [tex]\frac{hc}{λ}[/tex]
∴ E₁ = [tex]\frac{hc}{3λ}[/tex] ------ ( 2 )
Equate equations ( 1 ) and ( 2 )
[tex]\frac{h^{2} }{8mL^{2} }[/tex] = [tex]\frac{hc}{3λ}[/tex] ----- ( 3 )
where ; L = width of the box
Final step : Determine the width of the box
from equation ( 3 )
L² = [tex]\frac{3hλ }{8mc}[/tex] ---- ( 4 )
∴ L² = ( 6.6 * 10⁻³⁴ * 122 * 3 * 10⁻⁹ ) / ( 8 * 9.1 * 10⁻³¹ * 3 * 10⁸ )
∴ L = 3.3 * 10⁻¹⁰ m
Hence we can conclude that The width of the box in order for the n = 2 to n = 1 is : 3.3 * 10⁻¹⁰ m
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PART A If the atom is modeled as an electron in a one-dimensional box, what is the width of the box in order for the n = 2 to n = 1 transition to correspond to emission of a photon of this energy?
