View source

From WolfWikis

You do not have permission to do that, for the following reason:
The action you have requested is limited to users in the group user.
You can view and copy the source of this page:
{{nav|CH 101|Exercises |{{CH 101num_exerc}}}}
{{(}}
Back to [[CH 101]]
{{)}}

==Useful formulas and numbers==

:E<sub>'''n'''</sub>= - hR(Z/'''n''')<sup>2</sup>

:h= Planck's constant
:R= Rydberg constant
:hR= 2.18x10<sup>-18</sup> Joule
:Z= the charge of the nucleus = +1 for hydrogen.

:h=6.626x10<sup>-34</sup>Js
:e= 1.6021<sup>-19</sup>C
==Exercises==
Consider a single hydrogen atom in vacuum.

#Calculate the ionization (binding) energy of a H atom, when its electron is in the state '''n'''=2 and '''n'''=17.
#:<font color=blue>
#:This is easy: we just need to plug in into the general formula:
#:E<sub>'''2'''</sub>= - hR(Z/'''2''')<sup>2</sup>= -5.45x10<sup>-19</sup> Joule.
#:E<sub>'''17'''</sub>= - hR(Z/'''17''')<sup>2</sup>= -7.545x10<sup>-21</sup> Joule.
#:Often these energies are expressed in electronvolts (eV) by dividing by the electron charge of the electron e= 1.6021<sup>-19</sup>C. We get
#:E<sub>'''2'''</sub>= -3.40 eV
#:E<sub>'''17'''</sub>= -0.04 eV
#: As we shall see later this has a lot to do with the voltages of batteries. Obviously '''n'''=2 requires a decent jolt to ionize, '''n'''=17 much less so.
#:</font>
#The bottom of the vacuum continuum corresponds with '''n'''= infinity. Why?
#:<font color=blue>
#:Again: just plug in! The '''n''' is in the denominator, so if we let it go to infinity (take limit, remember?) we get E=0 and yes we had just taken that to be the bottom of the continuum.
#:</font>
#What is the wavelength of the photon that can provide just enough energy to bring an electron with '''n'''=4 to the bottom of the continuum?
#:<font color=blue>
#:This requires a bit more conversion, but let's start with the energy in Joules:
#:E<sub>'''4'''</sub>= - hR(Z/'''4''')<sup>2</sup>= -1.3625x10<sup>-19</sup> Joule.
#:All this energy has to come from the photon. Its energy is E=hν, so that we can calculate its frequency in s<sup>-1</sup>
#: 1.3625x10<sup>-19</sup> [J] = 6.626x10<sup>-34</sup>[Js].ν[s<sup>-1</sup>]
#: The frequency ν=2.06E+14 [s<sup>-1</sup>]
#: To go to wavelength we need the speed of light: c=νλ = 2.99E+8 [m/s]=2.06E+14[s<sup>-1</sup>].λ[m]
#: λ=1457x10<sup>-9</sup>[m]= 1457[nm]. A wavelength somewhere in the infrared part of the electromagnetic spectrum.
#:</font>
#A UV light with a wavelength of λ=250nm is aimed at a gas of hydrogen atoms. Their electrons are either in the states '''n'''=1, 2 or 3. The kinetic energy of the liberated electrons are measured. Which value(s) do you expect to find for E<sub>kinetic</sub>?
#:<font color=blue>
#:Looks like we need to go the other way. Lets first calculate the energy of the photons
#: E=hν= h.c/λ= 6.626x10<sup>-34</sup>*2.99x10<sup>8</sup>/250x10<sup>-9</sup>=7.95E-19 Joules or 4.96 eV.
#: That is more than E<sub>'''2'''</sub>= 3.40 eV, so electrons in that state should get ejected with a kinetic energy of 4.96-3.40=1.56 eV.
#: For the other states we must calculate the binding energy first (see above). We find:
#: E<sub>'''1'''</sub>= -13.61 eV
#: E<sub>'''3'''</sub>= -1.51 eV
#: The '''n'''=3 electrons will leave with even more gusto: their kinetic energy will be 4.96-1.51= 3.45eV
#: However those '''n'''=1 electrons cannot be ejected. We just do not have enough energy to sever their tether. That would require 13.61eV and we only have 4.96...
#:</font>
#What is the energy required to promote an electron from '''n'''=1 to '''n'''=2?
#:<font color=blue>
#: In this case we must use our formula twice and compute the '''''difference''''' in energy (ΔE= E<sub>final</sub>-E<sub>initial</sub>) between the two states.
#: ΔE=E<sub>'''2'''</sub>-E<sub>'''1'''</sub>= -3.40+13.61 = +10.21 eV.
#: (The sign convention is not the big message here, but in case you wonder: the '''+''' means: this is a bill you have pay. Currency: a photon of the right energy. You put one in, the atom gobbles it up. It is called: ''absorption''.)
#:</font>
#What is the wavelength of a photon emitted when an electron goes from '''n'''=3 to '''n'''=2?
#:<font color=blue>
#:This goes the other way:
#:ΔE=E<sub>'''2'''</sub>-E<sub>'''3'''</sub>= -3.40+1.51 = -1.11 eV.
#:The negative sign tells us the hydrogen is paying the bill. It is kind enough to send us a photon of 1.11eV
#:Its wavelength takes a bit of conversion.
#:1.11eV corresponds to ν=2.68E+14 [s<sup>-1</sup>] or λ= 1117 nm, which is in the near infrared (red starts at 650 or so.)
#:</font>
Templates used on this page:
Return to CH 101/Keys 11.
Retrieved from "http://wikis.lib.ncsu.edu/index.php/CH_101/Keys_11"
View source

From WolfWikis

Views

Personal tools

Navigation

Search

Toolbox

Wiki Stats
