A particle of mass m moves in a one-dimensional box of length L, with boundaries

Question

A particle of mass m moves in a one-dimensional box of length L, with boundaries at x = 0 nm and x = 5 nm. Thus, V (x) = 0 for 0 x 5 nm, and V (x) = elsewhere.

a) Can light excite a particle from its ground state to the fourth excited state? Mathematically support your answer.

b) If the optical transition in (a) is possible, what is the required wavelength of light that generates such a transition?

c) Can light excite a particle from its state n=4 to state n=5? Mathematically support your answer.

Explanation / Answer

A particle of mass m moves in a one-dimensional box of length L,
with boundaries at x = 0 nm and x = 5 nm.

Thus, V (x) = 0 for 0 x 5 nm, and V (x) = elsewhere.

En = n^2*h^2/(8*m*L^2)

dE = E2 - E1 = (n2^2-n1^2)*h^2/(8*m*L^2)
lambda = h*c/dE = h*c*(8*m*L^2)/[(n2^2-n1^2)*h^2] = 8*m*L^2*c/[h*(n2^2-n1^2)]
lambda = 8*m*L^2*c/[h*(n2^2-n1^2)]

h = 6.626E-34    J·s
c = 3.0E8 m/s    
length, L = 5 nm = 5.0E-9 m

lambda = 8*m*L^2*c/[h*(n2^2-n1^2)] = 8*m*(5.0E-9)^2*(3.0E8)/[(6.626E-34)*(n2^2-n1^2)]
lambda (meter) = 90.55E24*m/(n2^2-n1^2)
lambda (nm) = 90.55E33*m/(n2^2-n1^2)

If the particle is electron, then mass, m = 9.11E-31 kg
lambda (nm) = 90.55E33*(9.11E-31)/(n2^2-n1^2) = 8.25E4/(n2^2-n1^2)
Visible wavelength of light is from 400-700 nm.

a) Can light excite a particle from its ground state to the fourth excited state?
Mathematically support your answer.
n2 = 4
n1 = 1
lambda (nm) = 8.25E4/(4^2-1^2) = 5500 nm
Visible wavelength of light is from 400-700 nm.

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