(a) 159 nm
First of all, let's calculate the energy difference between the level E1 and E4:
[tex]\Delta E=E_4 -E_1 = -2.01\cdot 10^{-19}J-(-1.45\cdot 10^{-18} J)=1.25\cdot 10^{-18} J[/tex]
Now we know that this energy difference is related to the wavelength of the absorbed photon by
[tex]\Delta E=\frac{hc}{\lambda}[/tex]
where
[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant
[tex]c=3\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda[/tex] is the wavelength of the photon
Solving for [tex]\lambda[/tex], we find
[tex]\lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.25\cdot 10^{-18} J}=1.59\cdot 10^{-7} m = 159 nm[/tex]
b) 293 nm
As done in part a), let's calculate the energy difference between the level E2 and E3:
[tex]\Delta E=E_3 -E_2 = -5.71\cdot 10^{-19}J-(-1.25\cdot 10^{-18} J)=6.79\cdot 10^{-19} J[/tex]
this energy difference is related to the wavelength of the absorbed photon by
[tex]\Delta E=\frac{hc}{\lambda}[/tex]
Solving for [tex]\lambda[/tex] again, we find
[tex]\lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{6.79\cdot 10^{-18} J}=2.93\cdot 10^{-7} m = 293 nm[/tex]
c) 226 nm
As done in part a) and b), let's calculate the energy difference between the level E1 and E3:
[tex]\Delta E=E_3 -E_1 = -5.71\cdot 10^{-19}J-(-1.45\cdot 10^{-18} J)=8.79\cdot 10^{-19} J[/tex]
this energy difference is related to the wavelength of the emitted photon by
[tex]\Delta E=\frac{hc}{\lambda}[/tex]
Solving for [tex]\lambda[/tex] again, we find
[tex]\lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{8.79\cdot 10^{-18} J}=2.26\cdot 10^{-7} m = 226 nm[/tex]
