# How do you calculate the ionization energy of a hydrogen atom in its ground state?

##### 1 Answer

#### Answer:

#### Explanation:

**!! VERY LONG ANSWER !!**

Start by calculating the **wavelength** of the emission line that corresponds to an electron that undergoes a

This transition is part of the **Lyman series** and takes place in the *ultraviolet* part of the electromagnetic spectrum.

Your tool of choice here 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 thewavelengthof the emitted photon (in a vacuum)#R# is theRydberg constant, equal to#1.097 * 10^(7)# #"m"^(-1)# #n_1# represents theprincipal quantum numberof the orbital that islower in energy#n_2# represents theprincipal quantum numberof the orbital that ishigher in energy

In your case, you have

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

Now, you know that as the value of **increases**, the value of **decreases**. When

#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 case, will get you

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

#1/(lamda) = R#

Rearrange to solve for the wavelength

#lamda = 1/R#

Plug in the value you have for

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

Now, in order to find the energy that corresponds to this transition, calculate the **frequency**, *emitted* when this transition takes place by using the fact that wavelength and 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 speed of light in a vacuum, usually given as#3 * 10^8# #"m s"^(-1)#

Rearrange to solve 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 energy 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 theenergyof the photon#h# isPlanck's constant, equal to#6.626 * 10^(-34)"J s"#

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 remove the electron from the ground state of a hydrogen atom in the gaseous state and create a hydrogen ion, you need to supply

This means that for **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 energy required to remove **mole** of electrons from **mole** of hydrogen atoms in the gaseous state.

To convert the energy to *kilojoules per mole*, use the fact that **mole** of photons contains **photons** as given by Avogadro's constant.

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 can thus say that for **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

My guess would be that the difference between the two results was caused by the value I used for Avogadro's constant and by rounding.

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