The ionization energy of an atom can be measured by photo-electron spectroscopy, in which light of wavelength l is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron 1EK2 is measured by determining its velocity, v since EK = 1/2 mv2. The Ei is then calculated using the relationship that the energy of the inci-dent light equals the sum of Ei plus EK.
(a) What is the ionization energy of rubidium atoms in kilo-joules per mole if light with l = 58.4 nm produces elec-trons with a velocity of 2.450 * 106m/s? (The mass of an electron is 9.109 * 10-31 kg.)

Question

The ionization energy of an atom can be measured by photo-electron spectroscopy, in which light of wavelength l is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron 1EK2 is measured by determining its velocity, v since EK = 1/2 mv2. The Ei is then calculated using the relationship that the energy of the inci-dent light equals the sum of Ei plus EK. 
(a) What is the ionization energy of rubidium atoms in kilo-joules per mole if light with l = 58.4 nm produces elec-trons with a velocity of 2.450 * 106m/s? (The mass of an electron is 9.109 * 10-31 kg.)

Accepted Answer

Hey everyone, we're told that photoelectron spectroscopy pes can be used to measure the ionization energy of an atom. In this method, the atom is exposed to a beam of light with a wavelength ejecting an electron from the atom using its velocity and the equation kinetic energy equals half Mv squared The kinetic energy of the electron is determined. The ionization energy is then calculated from the relationship that the sum of the ionization energy and the kinetic energy is equal to the energy of the incident light. Using this information, calculate the ionization energy of a lithium atom. If light with a wavelength of 30.5 nanometers ejected electrons with a velocity of 3.5 to 3 times since the six m per second, The mass of an electron is 9.109 times 10 to the negative 31st kg. First let's go ahead and calculate the energy of our incident light. So we have energy equals our planes constant times the speed of light divided by our wavelength. So, plugging in these values, we know that our planks constant is 6.626 times to the negative 30 for jules times second. We're going to multiply this by our speed of light which is 3.00 times 10 to the eight m/s. Now let's go ahead and divide this by our wavelength, we were told that we had a wavelength of 30.5 nanometers. But first we need to convert this into meters in order to eliminate the meters in our numerator. Using our dimensional analysis, we know that we have 10 to the nine nanometers per one m. Now when we calculate this out and cancel out all of our units, We end up with an energy of 6.5174 times 10 to the -18 jewels. Now let's go ahead and calculate the energy of the electron and we can do so by using our formula kinetic energy equals half M V squared. Plugging in our values. We have the mass of our electron which was told to us to be 9.109 Times 10 to the -31 kg. And we were told that we had a velocity of 3.5-3 times 10 to the six m/s. And we square this. Now when we calculate this out, we end up with 5.65 to 8 times 10 to the negative 18 kg times meters squared over Second squared Now we've learned that one kg times meter squared over. Second squared is the same as jewels. So we can simply rewrite this as 5.65 to 8 times 10 to the negative 18 jewels. Now let's go ahead and calculate our ionization energy. We were told that our energy total is equivalent to our ionization energy plus our kinetic energy, solving for our ionization energy, we get our total energy minus our kinetic energy. So let's go ahead and plug in these values. We have 6.5174 times 10 to the negative 18 jewels, which was our total energy. Now let's go ahead and subtract our kinetic energy which we calculated to be 5.65 to 8 times 10 to the negative jewels. So when we calculate this out, we end up with an ionization energy of 8. times 10 to the -19 joules per photon. Now to get our ionization energy into kilo jewels, we're going to take our 8.646 times 10 to the negative jewels over photon. And we're going to use dimensional analysis to convert this. We know that we have tens of their jewels per one, kill a jewel And we know that one mole Contains 6.02, 2 times 10 to the 23rd photons. Now, when we calculate this out and cancel out all of our units, we end up with an ionization energy of 521 kg joules per mole. And this is going to be our final answer. Now, I hope that made sense. And let us know if you have any questions

Verified Solution

Key Concepts

Ionization Energy

Photoelectron Spectroscopy

Kinetic Energy of Electrons