Use Coulomb's law to calculate the ionization energy in kJ>mol of an atom composed of a proton and an electron separated by 100.00 pm. What wavelength of light has sufficient energy to ionize the atom?

Video transcript

welcome back everyone. One of the many applications of columns law is that it can be used to calculate ionization. In hopes of simple atoms calculate the ionization energy for a hydrogen atom take its atomic radius equal to 120 pick a meters and also determine the wavelength of light beam that could be used to ionized the hydrogen atom. So to solve a problem like this, we want to recall our columns law which is where we relate the potential energy of our atom two columns constant, which is going to be represented by the following quotient, where we take 1/4 times pi times epsilon not. And then this is going to be multiplied by the next quotient here in parentheses where we take Q. One, the charge of a electron times Q. To the charge of a proton divided by r the radius of our charged particles or between are charged particles. So we're going to make note of the fact that epsilon not is equal to a value of and sorry, this is an equal sun epsilon naught is equal to a value of 8.85 times 10 to the negative 12 power. And our units are Kalume square divided by jules times meters. So referring to the prompt were given our hydrogen atom and we want to recall that a hydrogen atom contains one proton and therefore one electron because it's in group one A. On the periodic table. And so that is why it corresponds to the following proton number and number of electrons since it's a neutral hydrogen atom. So just noting down are important details. This entire first quotient is columns constant. So we'll actually put these parentheses in red and then our again represents the distance Between Q one and Q two. Where we want to make note of the fact that Q1 again describes the charge of our electron. And we're going to recall that. That's a value of negative 1.60218 times 10 to the negative 19th power columns. And Q two describes our charge of the proton in hydrogen, which we should recall has a value of positive 1. oh 218 times 10 to the negative 19 power columns. Now recognize that given in the prompt we have a radius Given that is equal to 120 PK m. But we want to recognize that of course we use units of meters in our calculation for potential energy of an atom. We use meters here in part of our columns constant. So we're going to have to convert from PICO meters. two m. So pika meters would be in the denominator and meters in the numerator and recall that our prefix PICO tells us that for one pick a meter we have 10 to the negative 12 power of our base unit meter. So we can cancel out PICO meters were left with meters and this is going to give us a atomic radius equal to 1.20 times 10 to the negative 10th power meters. So now we can go into calculating the potential energy of our hydrogen atom where again we have 1/ or rather we want to carefully set it up so that it's one over four times pi times r value for epsilon not given as or which we recall as 8.85 times 10 to the negative 12 power column squared, divided by jules times meters. And actually we can go ahead and write in our value for pi. So let's recall that Pi is a value of just 3.14. So this completes our values for our first quotient, which is multiplied by we have Q1 and Q2 where Q1 we recall is negative 1.602 times 10. Or sorry, 1.602 18 times 10 to the negative 19th power columns, multiplied by We have positive one. We don't really need to write in the positive sign. So just 1.60218 times 10 to the negative 19th power columns for our proton charge. And then this numerator is divided by our denominator which is our radius. We converted to meters as 1.2 times to the negative 10th power meters. And so now we have our second quotient and we can just simplify carefully in our calculators to find that our potential energy of the hydrogen atom Is equal to the product between the value of our first quotient. Which should equal 8. times 10 to the 9th power. And our units are columns. Or rather we'll take care of units in one second. Let's just write out our value of our products here. So for our second quotient we're going to have a value of negative 2.1391 times 10 to the negative. And sorry, there's a five here. 1915 times 10 to the negative 28th power. And so when we write out our units, let's get rid of columns squared with columns squared in the numerator because they're going to be multiplied by one another here. We can also cancel out meters. And that leaves us with units of jewels for our first quotient. And so what will result in when we multiply the product here is the potential energy of our atom equal to a value of negative 1. 58 times 10 to the negative 18th power. And we have units of jewels which we're going to interpret per hydrogen atom. And so now that we have this potential energy, we want to get the ionization energy by converting this into killing joules per mole as well as by recalling our formula where ionization energy is found from taking the potential energy of an atom of an electron. Sorry, that is free from the atom. So that would be an electron with a potential energy of infinity which is subtracted from the potential energy of an electron that it's still within the atom in its shell. So recall that shell numbers or the energy level of a shell is represented by the term end. So we have sub end here. So taking these two steps are going to result in our ionization energy of our hydrogen atom. So let's begin with that calculation. We want to first take the potential energy of our electron that is free from the hydrogen atom. And so we want to recall that this value potential energy of electron free from adam is going to equal a value of zero. And that's because we recall that ionization energy is defined as energy removing an electron from its shell. So if the potential energy of an electron free from the atom is equal to zero, that's why we have this value of zero because the electron is already free from the atoms. So it wouldn't require any ionization energy to remove this electron. Now for our potential energy of our electron in its shell. So still within the atom. This is going to equal the potential energy of our atom that we just calculated above as negative 1.924458 times 10 to the negative 18th power jewels per atom. So we're going to calculate our organization energy by taking the difference between these two values. So we'll have zero minus negative 1.924458 times 10 to the negative 18th power jewels per atom. And so taking this difference, we're really going to be adding here. And so this means that our ionization energy is now equal to positive 1.924458 times 10 to the negative 19th power, sorry, times 10 to the negative 18th power jewels per adam. And again we need this to be in kill jules Permal. So we're now going to multiply by our conversion factor to get rid of the term atom. So we're going to utilize avocados number, Which we recall is 6.022 times 10 to the 23rd power. And we can use the interpretation of atoms now to get rid of atoms and we're going to say that this is equivalent to one mole. And now we want to get killed jules in the picture. So we're going to cancel out jewels by placing it in the denominator and kill a jewels in the numerator where we recall that are prefixed kilo tells us we have 10 to the third power of our base unit jewels. So canceling out jewels were left with kayla jules Permal, which is what we want for our ionization energy. And this is going to yield the result of 1.16 times 10 to the third power kilo, joules per mole. And make note that this is a positive value for our ionization energy. So this is going to be our first answer for our ionization energy of our hydrogen atom, which tells us the amount of energy required to remove an electron from this hydrogen atom. And now we need to find the wavelength of light that we could use in a beam to ionized the hydrogen atom, meaning we want to recall our formula which relates the energy of a photon of light. Two. Plank's constant represented by H, multiplied by our speed of light. C. Which is divided by lambda, which we recall is our wavelength term. And so solving for wavelength because we want that to answer this prompts, we're going to reorganize this so that we have planks constant times the speed of light now divided by the energy of our photon. And for our energy of our photon, we're going to say that that would be equivalent to our organization energy which above we calculated as 1.924458 times 10 to the negative 18th power jewels per atom. So we're going to use that same value. And so we'll have in our numerator plank's constant, which we recall is a value of 6.626 times 10 to the negative 34th power units of jewels times seconds, which is multiplied by our speed of light, which we want to recall as a value of 3.0 times 10 to the eighth power meters per second in our denominator. We're dividing by the energy of a photon of light which we're relating to. The ionization energy being what we calculated as 1.924458 times 10 to the negative 18th power jewels per hydrogen atom. And so simplifying this in our calculators Is going to yield a result of 1.03 times 10 to the negative seventh power. And as far as our units will be able to get rid of jewels as well as seconds. And we will be left with meters where Adam is just a interpreted unit. So we can technically cancel it out to. And so we have units of meters but we want to recall that our unit for wavelength is in nanometers typically. So we're going to convert from meters to nanometers in the numerator, where we would recall that our prefix nano tells us that we have 10 to the negative ninth power of our base unit meter. And so canceling out meters were left with nanometers as our final unit for lambda, which is going to result in a wavelength equal to a value of 103 nanometers. And this would be our second final answer as our wavelength of radiation. Two iron eyes, the hydrogen atom. So what's highlighted in yellow are our two final answers to complete this example. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.