The first ionization energy of sodium is 496 kJmol. Use Coulomb’...

Question

The first ionization energy of sodium is 496 kJ>mol. Use Coulomb’s 
law to estimate the average distance between the sodium 
nucleus and the 3s electron. How does this distance compare to 
the atomic radius of sodium? Explain the difference.

Accepted Answer

everyone in this example, we need to use columns law to find the theoretical atomic raid. I given a ionization energy or first ionization entropy of 520 kg joules per mole for a lithium atom. We also need to find the calculated value or compare the calculated value with the actual atomic radius of our lithium atom and describe why the two values are different. So we need to make note of in our textbooks the actual atomic radius of our lithium atom. And looking that up in our textbooks or online we would see that that radius is equal to a value of 167 km for the lithium atom. And so what we want to do next is also recall our formula for columns law which tells us that our potential energy of our atom or electrostatic potential energy E. Is equal to 1/4 times pi times E. Not which we recall is our columns constant. And then this quantity is then multiplied by the second quotient where we have the absolute value of our Q. One value which we recall is the charge of our electron here multiplied by our second particle particle here which is Q. To the charge of our nucleus. We end the absolute value animator. And then in the denominator we're going to be dividing by the radius which is the distance between our two charged particles which was our nucleus for Q two and our charge of our electron for Q. One. So what we wanna do is isolate this formula for columns law for radius. And what we would do is have radius equal 1/4 times pi still times our columns constant. N. Not now being multiplied by our electrostatic potential energy E. And then this quantity is then multiplied by the absolute value of Q. One times Q two and looking up our charge of an electron. We would recall that that value for Q one Is equal to a value of negative 1.60 times 10 to the negative 19th power columns per atom of our lithium. And then for our charge of our nucleus which is our value for Q two. We would recall that equal to a positive value 1.60 times 10 to the negative 19th power columns per lithium atom. So we can write this in with a plus in front to emphasize the fact that this is a positive value. So we want to plug in all of our known info into this formula. But before we do so we need to recognize that they give us our first ionization entropy In units of 520 kg joules per mole. And we want to turn this into our potential energy potential electrostatic energy which was our e. Variable here in units of jewels per atom. So what we want to do is start out with that given value there which they give us the electrostatic potential energy equal to 520 kg joules per mole for our lithium atom. And we're going to convert from kayla jewels two jewels first. And we want to recall that our prefix kilo tells us that we should have 10 to the third power jewels. So now we're able to cancel out our units of kilo jewels. Leaving us with jewels. And now from jules we want to end up with jules per adam. So we're going to recall that for one mole of our lithium atom. We have a total of six point oh 22 times 10 to the 23rd power atoms, which we recall is avocados number. And so now we're able to cancel out moles and moles, leaving us with our final units jewels per atom Which is our units for electrostatic potential energy. And what we would get is that our electrostatic potential energy is equal to a value of 8.64 times 10 to the negative 19th power jewels per atom. And so now we want to go ahead and plug this into our formula for the radius. And this is going to be our theoretical radius. That we're calculating here based on the given ionization entropy. So what we're going to get is r is equal to what we said for Formula 1/4 times pi times not which we should recall is our columns constant equal to a value of 8.85 times 10 to the negative 12 power columns or columns squared divided by jules times moles. So this is our first part of our equation. And then we're going to also extend this and multiply it by our electrostatic potential energy which we calculated above in jewels per atom. And we'll plug that in as 8.64 times 10 to the negative 19th power jewels per atom. And so this completes the first part of our formula here. So we'll just have one in the middle. So everything's evened out. And we're going to multiply this entire quotient by the absolute value of Q one times Q two. Now we should recall that the absolute value symbol essentially turns anything negative to being positive. So that would make both our charge of our electron and our charge of our nucleus Q one and Q two, both equal to a value of plus 1.60 times 10 to the negative 19th power columns. And so based on that information, we can go ahead and just multiply this by positive and I'm just gonna make sure we have enough room. We can multiply this by positive 1.60 times 10 to the negative 19th power columns per atom squared since they're both going to be the same value for both the charge of our electron in charge of our nucleus when we apply the absolute value simple. And so what we would get is that our radius is equal to when we take the value in our calculators for this first quantity here, we should get a value of 1.4072 times to the 28th power. And we're left with units of moles. And I'll show you how that canceled out because we can go ahead and get rid of our column squared with the column squared over here we can get rid of jewels with jewels and we can get rid of atoms with atoms leaving us with just moles. And so this is what we get for our first quotient there and then for the second product, which was our charge of our electron in charge of our nucleus squared. We should get a value of 2.56 times 10 to the negative 38th power. And so taking the two products here, we're going to get a radius value equal to 2.6624 times 10 to the negative 10th power. And units of meters however, we want our radius to be in units of PICO meters because that is what we're given for lithium atom. According to our text book value of the radius for lithium atom. And so we're going to do a conversion factor to go from meters to PICO meters by recalling that are prefixed PICO tells us we have 10 to the 12 power PICO meters for one m. And so this is going to give us a value after we cancel out the units of meters, We're going to get a value equal to 266 m. And as we can see our theoretical radius 266 PICO meters is greater than 167 PICO meters for the lithium atom. For the last part of our question, we need to explain why these values are different. And so this difference, I'm just gonna scroll down for more room is likely due to the electron removed from lithium configuration, which we should recall is going to be In the two S 1 sub shell. And so essentially because we only have this one electron in our two S sub shell for lithium, we would say this is therefore lowers the actual radius of lithium. And so to complete this example, this entire explanation here as well as our calculated value for our radius and PICO meters is going to be our final answer to complete this example. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Watch next