Ionization involves completely removing an electron from an atom....

Question

Ionization involves completely removing an electron from an atom.
How much energy is required to ionize a hydrogen atom in its
ground (or lowest energy) state? What wavelength of light contains
enough energy in a single photon to ionize a hydrogen atom?

Accepted Answer

Hey everyone in this example, we have free radical reactions which are abundant in organic reactions. And we're told that lights used to initiate this type of reaction by breaking bonds between the atoms to form radicals. We need to consider the bond and a chlorine molecule which takes about 239 kg joules of mole Permal of energy to break. And we need to calculate the longest possible wavelength needed to initiate this type of free radical reaction. So we're going to go ahead and call our formula for wavelength, which we should recall is represented by lambda and we should recall. That wavelength is found by taking plank's constant, Multiplying it by the speed of light and then dividing it by our energy value. We should recall that. Our units for wavelength should be in meters and our units for energy should be in jewels. So our first step is to recognize that they give us our energy that is required to break the bond in a chlorine molecule as 239 kg joules per mole. So we want to convert from killing joules per mole, two jewels. So we're going to recall our conversion factor where we want to first focus on getting rid of that unit in the denominator moles by recalling that we have for one more 6.02, 2 times 10 to the 23rd power molecules which represent avocados number. And so we're using avocados number as a conversion factor so that we can cancel out moles. And now we have units of kilo joules per molecules. But what we want is units of jewels as our final unit. So we're going to multiply by a third conversion factor. Where we're going to go from kilo jewels in our denominator. So it can cancel out two jewels in our numerator. And so we should recall that our prefix kilo tells the stuff for one kg jewel. We have 10 to the third power jewels. So now we're able to cancel out kilo jewels with kilo joules in the denominator, leaving us with jewels as our final unit for energy. And this is jules per molecule. And so what we would get is a value equal to 3.97 times 10 to the negative 19th power jewels as our unit of energy here. And this is in joules per molecule of cl two. So now that we have our unit of energy and the proper units we can go ahead and find our wavelength. So what we're going to do is again recall in the numerator plank's constant which is represented as 6.626 times 10 to the negative 34th power in units of jewels times seconds. And then we're going to multiply this by our variable C, which represents the speed of light, which we recall is 3.0 times 10 to the eighth power meters per second In our denominator. We want to plug in our energy which we just converted into the proper unit jewels. So we're going to go ahead and plug in 3.97 times 10 to the negative 19th power jewels. And so before we go ahead and calculate the value, we're going to cancel out our unit. So we can get rid of jules because it's in the numerator and the denominator as well as seconds with inverse seconds in our numerator. This leaves us with moles. Or sorry, meters as our final unit in our numerator. So what we're going to have is a value equal to five point oh one times 10 to the negative seventh power meters. And now that we have our wavelength in meters, let's go ahead and convert our wavelength from meters to units of nanometers. So we're going to have five point oh one times 10 to the negative seventh power meters. And we're going to recall that. We want to cancel out meters to get two nanometers. So we should recall that our prefix nano gives us 10 to the negative ninth power of our base unit meters. So this allows us to go ahead and cancel out our units of meters, leaving us with nanometers as our final unit for wavelength. And this gives us a value equal to 501 nanometers as our longest possible wavelength needed to initiate the reaction to break the bond in our cl two molecule. So this would be our final answer to complete this example corresponding to choice D in our multiple choice. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

Watch next