Question

Calculate in kilojoules per mole the energy necessary to completely remove an electron from the first shell of a hydrogen atom (Rq = 1.097 * 10-2 nm-1).?

Answer 1

`E=1320"kJ/mol"`

Explanation:

We can use the Rydberg Expression to find the energy needed to move an electron from the `n=1` energy level to the `n=oo` energy level which means the atom is effectively ionised.

`1/lambda=R[(1)/(n_(1)^(2))-(1)/(n_2^2)]`

Where `n_1`< `n_2`

In this case `n_2=oo` so this becomes:

`1/lambda=R/1^2=R`

`:.lambda=I/R`

We need to correct the units of the Rydberg Constant to `m^(-1)rArr`

`R=1.097xx10^(7)m^(-1)`

So the wavelength of the photon required to eject the electron will be given by:

`lambda=(1)/(1.097xx10^(-7))=9.1157xx10^(-8)"m"`

The energy of the photon is given by:

`E=hf=(hc)/lambda=(6.626xx10^(-34)xx3xx10^(8))/(9.1157xx10^(-8))=2.180xx10^(-18)"J"`

This is the energy required to ionise a single atom. To find the energy required to ionise 1 mole of atoms we need to x by the Avogadro Constant:

`E=2.180xx10^(-18)xx6.02xx10^(23)=13.1xx10^(5)"J/mol"`

Convert to `"kJ"` by dividing by 1000 `rArr`

`E=1310"kJ/mol"`

Answer 2

`2.181 * 10^(-21)"kJ/atom"" "`, or `" ""1313 kJ/mol"`

Explanation:

The idea here is to use Rydberg's equation to find the wavelength of the emitted electromagnetic radiation first, then convert this wavelength to energy using the Einstein-Planck equation.

So, the Rydberg equation looks like this

`1/(lamda) = R * (1/n_1^2 - 1/n_2^2)" "`, where

`lamda` - the wavelength of the emitted photon;
`R` - the Rydberg constant, equal to `1.097 * 10^(-2)"nm"^(-1)`;
`n_1` - the principal quantum number of the orbital from which the transition is taking place;
`n_2` - the principal quantum number of the orbital to which the transition is taking place.

Now, the first ionization energy is the energy needed to completely remove one mole of electrons from one mole of atoms in the gaseous state.

Since you didn't specify if you're looking for ionization energy per mole, I'll show you the value you'd have per atom.

So, completely removing an electron from an atom is equivalent to having `n_2 = + oo`.

Removing an electron from `n_1 = 1` to `n_2 = +oo` will require electromagnetic radiation of the wavelength

`1/(lamda) = R * (1/1^2 - 1/(oo)) = R * (1 - 0) = R`

This means that you have

`lamda = 1/R = 1/(1.097 * 10^(-2)"nm"^(-1)) = "91.16 nm"`

The Einstein-Planck equation, which establishes a relationship between photon energy and its frequency, looks like this

`E = h * f" "`, where

`R` - the energy of the photon;
`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`;
`f` - the frequency of the photon.

SInce frequency and wavelength have the following relationship

`c = lamda * f" "`, where

`c` - the speed of light in vaccuum, `~~ 3.0 * 10^(8)"ms"^(-1)`

it follows that you can write

`E = (h * c)/(lamda)`

Before doing any calculation, convert the wavelength from nanometers to meters

`91.16color(red)(cancel(color(black)("nm"))) * "1 m"/(10^(9)color(red)(cancel(color(black)("nm")))) = 9.116 * 10^(-8)"m"`

This means that you have

`E = (6.626 * 10^(-34)"J"color(red)(cancel(color(black)("s"))) * 3.0 * 10^(8)color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-1)))))/(9.116 * 10^(-8)color(red)(cancel(color(black)("m")))) = 2.181 * 10^(-18)"J"`

Expressed in kilojoules, this is equaivalent to

`2.181 * 10^(-18)color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = 2.181 * 10^(-21)"kJ"`

This means that if you supply this much energy to a hydrogen atom in its ground state, you will remove its electron completely.

SIDE NOTE To get the value in kJ per mole, the way it's usually reported, you need to first convert Joules to kilojoules, then use Avogadro's number.

`2.181 * 10^(-21)"kJ"/color(red)(cancel(color(black)("atom"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("atoms"))))/"mole" = "1313 kJ/mol"`