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

This change is component of the Lyguy series and takes location in the ultraviolet part of the electromagnetic spectrum. Your tool of alternative right here will be the Rydberg equation for the hydrogen atom, which looks prefer 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 major quantum number of the orbital that is lower in energy#n_2# represents the major quantum number of the orbital that is better in energy

In your case, you have actually

#(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 can say that

#1/n_2^2 -> 0#

This implies 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 certainly gain you

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

#1/(lamda) = R#

Rearrange to solve for the wavelength

#lamda = 1/R#

Plug in the worth 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 discover the power that synchronizes to this shift, calculate the frequency, #nu#, of a photon that is emitted when this shift takes location by making use of the truth that wavelength and also frequency have 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 rate of light in a vacuum, typically provided as #3 * 10^8# #"m s"^(-1)#

Rearvariety to solve for the frequency and plug in your worth to find

#nu * lamda = c indicates 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 straight 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 value 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 means that in order to rerelocate the electron from the ground state of a hydrogen atom in the gaseous state and develop a hydrogen ion, you must supply #2.181 * 10^(-18)# #"J"# of energy.

This suggests 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 forced to remove #1# mole of electrons from #1# mole of hydrogen atoms in the gaseous state.

To convert the energy to kilojoules per mole, use the truth that #1# mole of pholoads contains #6.022 * 10^(23)# photons as offered by Avogadro"s consistent.

You will 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 have the right to hence 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 energy of hydrogen is actually #"1312 kJ mol"^(-1)#.

You are watching: What wavelength of light contains enough energy in a single photon to ionize a hydrogen atom

See more: Living Off The Fat Of The Land "? Definition Of Live Off/On The Fat Of The Land My guess would certainly be that the difference in between the two outcomes was brought about by the worth I supplied for Avogadro"s constant and by rounding.

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