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)#