Start by calculating the wavelength of the emission line that corresponds to an electron that undergoes a #n=1 -> n = oo# transition in a hydrogen atom.

This shift is component of the Lymale series and also takes place in the ultraviolet component of the electromagnetic spectrum.

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Your tool of alternative below will be the Rydberg equation for the hydrogen atom, which looks like this

#1/(lamda_"e") = R * (1/n_1^2 - 1/n_2^2)#

Here

#lamda_"e"# is the wavelength of the emitted photon (in a vacuum)#R# is the Rydberg constant, equal to #1.097 * 10^(7)# #"m"^(-1)##n_1# represents the primary quantum number of the orbital that is reduced in energy#n_2# represents the principal quantum number of the orbital that is higher in energy

In your situation, you have

#(n_1 = 1), (n_2 = oo) :#

Now, you recognize that as the value of #n_2# increases, the value of #1/n_2^2# decreases. When #n=oo#, you deserve to say that

#1/n_2^2 -> 0#

This suggests that the Rydberg equation will take the form

#1/(lamda) = R * (1/n_1^2 - 0)#

#1/(lamda) = R * 1/n_1^2#

which, in your instance, will gain you

#1/(lamda) = R * 1/1^2#

#1/(lamda) = R#

Rearvariety to deal with for the wavelength

#lamda = 1/R#

Plug in the value you have actually for #R# to get

#lamda = 1/(1.097 * 10^(7)color(white)(.)"m") = 9.116 * 10^(-8)# #"m"#

Now, in order to find the power that corresponds to this change, calculate the frequency, #nu#, of a photon that is emitted as soon as this change takes location by making use of the truth that wavesize and also frequency have actually an inverse relationship described by this equation

#color(blue)(ul(color(black)(nu * lamda = c)))#

Here

#nu# is the frequency of the photon#c# is the speed of light in a vacuum, commonly provided as #3 * 10^8# #"m s"^(-1)#

Rearvariety to resolve for the frequency and plug in your value to find

#nu * lamda = c implies nu = c/(lamda)#

#nu = (3 * 10^(8) color(red)(cancel(color(black)("m"))) "s"^(-1))/(9.116 * 10^(-8)color(red)(cancel(color(black)("m")))) = 3.291 * 10^(15)# #"s"^(-1)#

Finally, the power of this photon is directly proportional to its frequency as described by the Planck - Einstein relation

#color(blue)(ul(color(black)(E = h * nu)))#

Here

#E# is the energy of the photon

Plug in your worth to find

#E = 6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 3.291 * 10^(15) color(red)(cancel(color(black)("s"^(-1))))#

#E = 2.181 * 10^(-18)# #"J"#

This indicates that in order to rerelocate the electron from the ground state of a hydrogen atom in the gaseous state and also create a hydrogen ion, you need to supply #2.181 * 10^(-18)# #"J"# of energy.

This means that for #1# atom of hydrogen in the gaseous state, you have

#"H"_ ((g)) + 2.181 * 10^(-18)color(white)(.)"J" -> "H"_ ((g))^(+) + "e"^(-)#

Now, the ionization energy of hydrogen represents the power compelled to remove #1# mole of electrons from #1# mole of hydrogen atoms in the gaseous state.

To convert the energy to kilojoules per mole, usage the fact that #1# mole of pholoads consists of #6.022 * 10^(23)# photons as given by Avogadro"s consistent.

You will certainly end up with

#6.022 * 10^(23) color(red)(cancel(color(black)("photons")))/"1 mole photons" * (2.181 * 10^(-18)color(white)(.)color(red)(cancel(color(black)("J"))))/(1color(red)(cancel(color(black)("photon")))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J"))))#

# = color(darkgreen)(ul(color(black)("1313 kJ mol"^(-1))))#

You can for this reason say that for #1# mole of hydrogen atoms in the gaseous state, you have

#"H"_ ((g)) + "1313 kJ" -> "H"_((g))^(+) + "e"^(-)#

The cited value for the ionization power of hydrogen is actually #"1312 kJ mol"^(-1)#.


You are watching: How much energy is required to ionize a hydrogen atom in its ground (or lowest energy) state?


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My guess would certainly be that the difference between the 2 results was resulted in by the value I used for Avogadro"s consistent and also by rounding.

#6.02 * 10^(23) -> "1312 kJ mol"^(-1)" vs "6.022 * 10^(23) -> "1313 kJ mol"^(-1)#