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Photoelectron spectroscopy applies the principle of the photoelectric effect to study...

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xkat123 | Student, Undergraduate | Honors

Posted July 14, 2013 at 5:00 AM via web

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Photoelectron spectroscopy applies the principle of the photoelectric effect to study orbital energies of atoms and molecules. High-energy radiation (usually UV or x-ray) is absorbed by a sample and an electron is ejected. The orbital enegy can be calculated from the known energy of the radiation and the measured energy of the electron lost. The following energy differences were determined for several electrons transitions:

change of energy from 2 to 1= 4.098x10^-17 J

change of energy from 3 to 1= 4.854x10^-17 J

change of energy from 5 to 1= 5.254x10^-17 J

change of energy from 4 to 2 = 1.024x10^-17 J

 

Calculate the energy change and the wavelength of a photon emitted in the following transition: level 5 to 4

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llltkl | College Teacher | Valedictorian

Posted July 14, 2013 at 5:40 AM (Answer #1)

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The atomic orbitals possess discrete energy levels. Transition of an electron from one level to another, emits (or absorbs) energy which is equivalent to the difference between the correwsponding energy levels,

i.e. `(E_i-E_f)=DeltaE=hnu=(hc)/lambda`

For the given atom under photoelectron spectroscopic investigation,

`E_2-E_1= 4.098*10^-17 J` ---- (i)

`E_3-E_1= 4.854*10^-17 J` ---- (ii)

`E_5-E_1= 5.254*10^-17 J` ---- (iii)

`E_4-E_2= 1.024*10^-17 J` ---- (iv)

-----------------------------------------

(iii)-{(iv)+(i)}

`E_5-E_1-E_4+E_2-E_2+E_1=5.254*10^-17-{1.024*10^-17 J+4.098*10^-17 J}`

`rArr E_5-E_4=0.132*10^-17 J`

`=1.32*10^-18 J`

Wavelength of the associated radiation is given by,

`lambda=(h*c)/(DeltaE)`

`=(6.626*10^-34 * 2.998*10^8)/(1.32*10^-18) m`

`=1.505*10^-7 m`

`=150.5 nm`

Therefore, in the transition from level 5 to level 4, the energy change is 1.32*10^-18 J and wavelength of the associated photon is 150.5 nm.

 

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