Electronic transitions (i. e., absorption of uv or visible light) of the conjugated molecule butadiene can be approximated using the particle-on-a-line model, assuming one of the pi electrons can move along the whole four-carbon chain. A molecule of butadiene is found to absorb light of 207 nm corresponding to the electron being excited from the n=2 to n=3 levels. Use this to determine the length, in Angstroms, of the four-carbon chain to which the electron is confined.

2.51 Angstroms

Explanation:

For a particle in a one dimensional box, the energy level, En, is given by the expression:

En = n²π² ħ² / 2ma²

where n is the energy level, ħ²  is Planck constant divided into 2π, m is the mass of the electron ( 9.1  x 10⁻³¹ Kg ), and a is the length of the one dimensional box.

We can calculate  the change in energy, ΔE, from n = 2 to n= 3  since we know the wavelength of the transition  ( ΔE = h c/λ ) and then substitute this value for the expresion of the ΔE for a particle in a box and solve for the  length a.

λ = 207 nm x 1 x 10⁻⁹ m/nm = 2.07 x 10⁻⁷ m      ( SI units )

ΔE = 6.626 x 10⁻³⁴ J·s x  3 x 10⁸ m/s  / 2.07 x 10⁻⁷ m

ΔE   = 9.60 x 10⁻¹⁹ J

ΔE(2⇒3) = ( 3 - 2 )  x  π² x ( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )

9.60 x 10⁻¹⁹ J  =  π² x( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )

⇒ a = 2.51 x 10⁻¹⁰ m

Converting to Angstroms:

a = 2.51 x 10⁻¹⁰ m x 1 x 10¹⁰ Angstrom / m = 2.51 Angstroms

thats a little if u want the answer just look it up on google thats what i do haha : )