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Calculate the wavelength of the light emitted when an electron in a one-dimensional box of length 5.4 nmnm makes a transition from the n=9n=9 state to the n=8n=8 state.

Answer :

Answer : The wavelength of the light emitted is, [tex]5.66\times 10^{-24}m[/tex]

Explanation :

The energy level of quantum particle in a one-dimensional box is given as:

[tex]E_n=\frac{n^2h^2}{8mL^2}[/tex]

or,

[tex]\Delta E=E_9-E_8=\frac{n_9^2h^2}{8mL^2}-\frac{n_8^2h^2}{8mL^2}[/tex]

[tex]\Delta E=E_9-E_8=\frac{h^2}{8mL^2}\times (n_9^2-n_8^2)[/tex]

where,

[tex]E_n[/tex] = change in energy

n = energy level

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

m = mass of electron = [tex]9.109\times 10^{-31}kg[/tex]

L = length of a one-dimensional box = [tex]5.4nm=5.4\times 10^{-9}m[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta E=E_9-E_8=\frac{(6.626\times 10^{-34}Js)^2}{8\times (9.109\times 10^{-31}kg)\times (5.4\times 10^{-9}m)^2}\times [(9)^2-(8)^2][/tex]

[tex]\Delta E=E_9-E_8=3.5124\times 10^{-2}J[/tex]

Now we have to calculate the wavelength of the light emitted.

[tex]\Delta E=\frac{hc}{\lambda}[/tex]

where,

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light = [tex]3\times 10^{8}m/s[/tex]

[tex]\lambda[/tex] = wavelength of the light

Now put all the given values in the above formula, we get:

[tex]3.5124\times 10^{-2}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^{8}m/s)}{\lambda}[/tex]

[tex]\lambda=5.66\times 10^{-24}m[/tex]

Thus, the wavelength of the light emitted is, [tex]5.66\times 10^{-24}m[/tex]

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