CH 101/Keys 11

From WolfWikis

Useful formulas and numbers

E_n= - hR(Z/n)²

h= Planck's constant

R= Rydberg constant

hR= 2.18x10^-18 Joule

Z= the charge of the nucleus = +1 for hydrogen.

h=6.626x10^-34Js

e= 1.6021^-19C

Exercises

Consider a single hydrogen atom in vacuum.

1. Calculate the ionization (binding) energy of a H atom, when its electron is in the state n=2 and n=17.

   This is easy: we just need to plug in into the general formula:

   E₂= - hR(Z/2)²= -5.45x10^-19 Joule.

   E₁₇= - hR(Z/17)²= -7.545x10^-21 Joule.

   Often these energies are expressed in electronvolts (eV) by dividing by the electron charge of the electron e= 1.6021^-19C. We get

   E₂= -3.40 eV

   E₁₇= -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.

2. The bottom of the vacuum continuum corresponds with n= infinity. Why?

   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.

3. 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?

   This requires a bit more conversion, but let's start with the energy in Joules:

   E₄= - hR(Z/4)²= -1.3625x10^-19 Joule.

   All this energy has to come from the photon. Its energy is E=hν, so that we can calculate its frequency in s^-1

   1.3625x10^-19 [J] = 6.626x10^-34[Js].ν[s^-1]

   The frequency ν=2.06E+14 [s^-1]

   To go to wavelength we need the speed of light: c=νλ = 2.99E+8 [m/s]=2.06E+14[s^-1].λ[m]

   λ=1457x10^-9[m]= 1457[nm]. A wavelength somewhere in the infrared part of the electromagnetic spectrum.

4. 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_kinetic?

   Looks like we need to go the other way. Lets first calculate the energy of the photons

   E=hν= h.c/λ= 6.626x10^-34*2.99x10⁸/250x10^-9=7.95E-19 Joules or 4.96 eV.

   That is more than E₂= 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:

   E₁= -13.61 eV

   E₃= -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...

5. What is the energy required to promote an electron from n=1 to n=2?

   In this case we must use our formula twice and compute the difference in energy (ΔE= E_final-E_initial) between the two states.

   ΔE=E₂-E₁= -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.)

6. What is the wavelength of a photon emitted when an electron goes from n=3 to n=2?

   This goes the other way:

   ΔE=E₂-E₃= -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^-1] or λ= 1117 nm, which is in the near infrared (red starts at 650 or so.)