**!! VERY LONG ANSWER !!**
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 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 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
**principal quantum number** of the orbital that is **lower in energy**
- #n_2# represents the
**principal quantum number** of the orbital that is **higher in energy**
In your case, you have
#{(n_1 = 1), (n_2 = oo) :}#
Now, you know 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 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 #R# to get
#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**, #nu#, of a photon that is 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 the
**energy** of the photon
- #h# is
**Planck'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 #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 energy required 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 fact that #1# **mole** of photons contains #6.022 * 10^(23)# **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 #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)#.
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)#
$\begingroup$ What energy would be needed to remove the electron from the $n = 4$ level of the hydrogen atom? - $\pu{−3.49 * 10^{−17} J}$
- $\pu{−1.36 * 10^{−19} J}$
- $\pu{+2.18 * 10^{−18} J}$
- $\pu{+1.36 * 10^{−19} J}$
I assume that the way to do this is to start from the Rydberg formula, $$
E = \mathcal{R}Z^2 \left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right),
$$ set the initial level of the electron as $n_i = 4$, and the final level corresponding to removing the electron (ionization) as $n_f = \infty \implies \frac{1}{\infty^2} = 0$, leading to $$
E = \pu{13.6*\frac{1}{4^2} eV = 0.85 eV = 1.36*10^{-19} J}.
$$ $\endgroup$ |