Answer:
The edge of the length is [tex]\mathbf{L = 8.54 \times 10^{-10} \ m}[/tex]
Explanation:
From the given information:
The associated energy for a particle in three - dimensional box can be expressed as:
[tex]E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)[/tex]
here;
h = planck's constant = [tex]6.626 \times 10^{-34} \ Js[/tex]
[tex]n_i[/tex] = the quantum no in a specified direction
m = mass (of particle)
L = length of the box
At the ground state [tex]n_x = n_y = n_z=1[/tex]
The energy at the ground state can be calculated by using the formula:
[tex]E_1 =\dfrac{3h^2}{8mL^2}[/tex]
At first excited energy level, one of the quantum values will be 2 and the others will be 1.
Thus, the first excited energy will be: 2,1,1
∴
[tex]E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}[/tex]
[tex]E_2 =\dfrac{(4+1+1)h^2}{8mL^2}[/tex]
[tex]E_2 =\dfrac{(6)h^2}{8mL^2}[/tex]
The transition energy needed to move from the ground to the excited state is:
[tex]\Delta E= E_2 - E_1[/tex]
[tex]\Delta E= \dfrac{6h^2}{8mL^2}- \dfrac{3h^2}{8mL^2}[/tex]
[tex]\Delta E= \dfrac{3h^2}{8mL^2}}[/tex] ----- (1)
Recall that:
the wavelength identified with the electronic transition is: 800 nm
800 nm = 8.0 × 10⁻⁷ m
However, the energy-related with the electronic transition is:
[tex]\Delta E =\dfrac{hc}{\lambda}[/tex]
[tex]\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }[/tex]
[tex]\Delta E =2.48 \times 10^{-19} \ J[/tex]
Replacing the value of [tex]\Delta E[/tex] in (1); then:
[tex]2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}[/tex]
Making the edge length L the subject of the formula; we have:
[tex]L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }[/tex]
[tex]L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }[/tex]
[tex]\mathbf{L = 8.54 \times 10^{-10} \ m}[/tex]
Thus, the edge of the length is [tex]\mathbf{L = 8.54 \times 10^{-10} \ m}[/tex]
