What is the ionization energy of rubidium atoms (in kilojoules per mole) if light with λ = 58.4 nm produces electrons with a velocity of 2.450×10 6 m/s 2.450*10^6 m/s?

1 Answer
Jun 25, 2016

Answer:

#402.9" ""kJ/mol"#

Explanation:

The first ionisation energy is the energy required to remove 1 mole of electrons from 1 mole of atoms in the gaseous state.

It is the enthalpy change for:

#X_((g))rarrX_((g))^(+)+e#

In this example I will assume that we have gaseous rubidium atoms.

When a photon strikes a rubidium atom it may have sufficient energy to excite electrons to higher energy levels:

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If the photon has a lot of energy it may cause the electron to be ejected from the atom forming an ion.

The excess energy of the photon, after ionisation, will appear as the kinetic energy of the ejected electron.

We can use the conservation of energy to set this up as an equation:

Photon energy = ionisation energy + kinetic energy of electron

The energy of the photon is given by the Planck expression:

#E=hf#

#f# is the frequency of the photon.

#h# is the Planck Constant with the value of #6.626xx10^(-34)" " "Js"#

The kinetic energy of the electron is given by:

#"KE"=1/2mv^2#

#m# is the mass #=9.11xx10^(-31)" ""kg"#

#v# is the velocity #=2.45xx10^(6)" ""m/s"#

So we can say:

#hf="IE"+1/2mv^(2)#

#:."IE"=hf-1/2mv^2#

Since #c=fxxlambda# we can write this as:

#"IE"=(hc)/lambda-1/2mv^2#

Putting in the numbers:

#"IE"=(6.626xx10^(-34)xx3xx10^(8))/(58.4xx10^(-9))-[1/2xx9.11xx10^(-31)xx(2.45xx10^6)^2]#

#"IE"=[3.403xx10^(-18)]-[2.734xx10^(-18)]" ""J"#

#"IE"=6.690xx10^(-19)" ""J"#

This is the energy required to ionise 1 atom of #"Rb"#

To get the energy required to ionise a mole of atoms we need to multiply this by the Avogadro Constant:

#"IE"=6.690xx10^(-19)xx6.022xx10^(23)=4.0287xx10^(5)" ""J/mol"#

Since there are 1000J in 1kJ this becomes:

#"IE"=402.9" ""kJ/mol"#

The literature value gives #403.032" ""kJ/mol"# so this seems a reasonable answer.