1 Answer

Daniel Broughan · Answer 1 · 2020-08-15T18:11:08+0000

(a) 159 nm

First of all, let's calculate the energy difference between the level E1 and E4:

$\Delta E=E_4 -E_1 = -2.01\cdot 10^(-19)J-(-1.45\cdot 10^(-18) J)=1.25\cdot 10^(-18) J$

Now we know that this energy difference is related to the wavelength of the absorbed photon by

$\Delta E=(hc)/(\lambda)$

where

$h=6.63\cdot 10^(-34) Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the photon

Solving for
$\lambda$ , we find

$\lambda=(hc)/(\Delta E)=((6.63\cdot 10^(-34) Js)(3\cdot 10^8 m/s))/(1.25\cdot 10^(-18) J)=1.59\cdot 10^(-7) m = 159 nm$

b) 293 nm

As done in part a), let's calculate the energy difference between the level E2 and E3:

$\Delta E=E_3 -E_2 = -5.71\cdot 10^(-19)J-(-1.25\cdot 10^(-18) J)=6.79\cdot 10^(-19) J$

this energy difference is related to the wavelength of the absorbed photon by

$\Delta E=(hc)/(\lambda)$

Solving for
$\lambda$ again, we find

$\lambda=(hc)/(\Delta E)=((6.63\cdot 10^(-34) Js)(3\cdot 10^8 m/s))/(6.79\cdot 10^(-18) J)=2.93\cdot 10^(-7) m = 293 nm$

c) 226 nm

As done in part a) and b), let's calculate the energy difference between the level E1 and E3:

$\Delta E=E_3 -E_1 = -5.71\cdot 10^(-19)J-(-1.45\cdot 10^(-18) J)=8.79\cdot 10^(-19) J$

this energy difference is related to the wavelength of the emitted photon by

$\Delta E=(hc)/(\lambda)$

Solving for
$\lambda$ again, we find

$\lambda=(hc)/(\Delta E)=((6.63\cdot 10^(-34) Js)(3\cdot 10^8 m/s))/(8.79\cdot 10^(-18) J)=2.26\cdot 10^(-7) m = 226 nm$

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