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.

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)#