Answer to Question #4312 in Physical Chemistry for Amanda
hc/l = DE = (h2/8ma2)[(k+1)2-k2] = (h2/8ma2)(2k+1)
Substituting for a,
DE = [h2/8m(1.4Å)2][(2k+1)/(k+2)2].
Solving for l,
l = (8m(1.4Å)2/h2)(hc)[(k+2)2/(2k+1)] = [(8m*(1.4x10-10 m)*c)/h][(k+2)2/(2k+1)].
The pertinent mass is the mass of an electron, 9.11x10-31 kg; c=3.00x108 m s-1; h = 6.63x10-34 J/s.
l = [(8*9.11x10-31 kg*(1.4x10-10 m)2*3.00x108 m s-1)/6.63x10-34 Js][(k+2)2/(2k+1)] = (6.46x10-8 m)[(k+2)2/(2k+1)] = (64.6 nm)[(k+2)2/(2k+1)].
Using guess and check methodology,
k l л/nm
The structure must have 15 double bonds: CH2=CH-(CH=CH)13-CH=CH2
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