Question

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 1

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

Explanation :

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

$E_n=\frac{n^2h^2}{8mL^2}$

or,

$\Delta E=E_9-E_8=\frac{n_9^2h^2}{8mL^2}-\frac{n_8^2h^2}{8mL^2}$

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

where,

$E_n$ = change in energy

n = energy level

h = Planck's constant = $6.626\times 10^{-34}Js$

m = mass of electron = $9.109\times 10^{-31}kg$

L = length of a one-dimensional box = $5.4nm=5.4\times 10^{-9}m$

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

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

$\Delta E=E_9-E_8=3.5124\times 10^{-2}J$

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

$\Delta E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

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

$\lambda$ = wavelength of the light

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

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

$\lambda=5.66\times 10^{-24}m$

Thus, the wavelength of the light emitted is, $5.66\times 10^{-24}m$