Answer:
Explanation:
The formula for determining the energy of state [tex]m_l[/tex] can be computed by using the formula:
[tex]Em_i = \dfrac{m_1^2h^2}{2I}[/tex]
Also, the momentum is:
[tex]l_2 = m_i h[/tex]
There are 22 electrons with two electrons in each of the lowest II energy levels so that the highest occupied states are [tex]m_1 = \pm 5[/tex]
The moment of inertia of an electron on a ring of radius 440 ppm is:
[tex]I = mR^2 \\ \\ = 9.109 \times 10^{31} \ kg (440 \times 10^{-12}) ^2 \\ \\ = 1.76 \times 10^{-49} \ kgm^2[/tex]
[tex]E_{\pm 5 } = \dfrac{25h^2}{2I} \\ \\ = \mathbf{7.89 \times 10^{-19} \ J}[/tex]
The angular momentum is:
[tex]l_2 = \pm 5h \\ \\ = \pm 5 \tiems ( 1.05457 \times 10^{-34} \ Js)[/tex]
[tex]= \mathbf{5.275 \times 10^{-34} \ Js}[/tex]
B) Let's recall that:
The lowest occupied energy level is [tex]m_1 = \pm 6[/tex] which implies that the energy [tex]E_{\pm 6}= 1.14 \times 10^{-18} \ J[/tex]
Thus;
[tex]\Delta E = E_{\pm 6} - E_{\pm 5} \\ \\ = 1.14 \times 10^{-18 \ J} - 0.79 \times 10^{-18} \ J \\ \\ = 0.35 \times 10^{-18} \ J \\ \\ 0.35 \times 10^{-18} \ J = hv \\ \\ 0.35 \times 10^{-18} \ J = h \dfrac{c}{\lambda}[/tex]
Hence, the radiation which would induce a transition that relates to the wavelength of about 570nm, a wavelength of visible light.
