Moseley established the concept of atomic number by studying X rays emitted by the elements. The X rays emitted by some of the elements have the following wavelengths: Element Wavelength (pm) Ne 1461 Ca 335.8 Zn 143.5 Zr 78.6 Sn 49.1 (e) A particular element emits X rays with a wavelength of 98.0 pm. What element do you think it is?

Video transcript

Hey everyone in this example, we're told that heavy elements emit characteristics of X rays when their electrons undergo transitions were shown below the wavelength of X rays emitted by these particular elements. And we're told that if a particular element emits an X ray with a wavelength of 1.662 extremes, we need to determine the identity of this element according to this prompt. We're also going to be plotting our square root of the frequency which we recall is represented by this symbol here versus atomic number, which we recall is represented by the symbol Z for our element. So what we should recognize is that we're given wavelength again in units of extremes recall that wavelength is represented by the symbol lambda and it's usually given in units of meters. So we're going to convert from angstrom is two m By recalling the conversion factor that for one angstrom, that is equivalent to 10 to the negative 10th power meters. So all we have to do to convert these wavelengths here is multiply by 10 to the negative 10th power. And then just put the unit meters. So we'll just do that for all of the wavelengths for our elements. And then for the last one for the element molybdenum, we're going to multiply by 10 to the negative 10th power. And sorry about that. That is going to give us our wavelength in meters equal to 7.107 times 10 to the negative 11th power. And we have units of meters. So now we have all of our wavelengths converted to meters. And what we should recognize is that because we need to plot the square root of our frequency, we don't have frequency. And so we're going to recall the formula to solve for frequency by utilizing the following formula. So recall that frequency is going to be found by taking our speed of light and dividing it by our wavelength lambda. And we should recall that frequency is typically expressed in units of hertz. So we should recall that for our units of the speed of light. That is equal to 3.0 times 10 to the eighth power meters per second. And let's just make this clear. And then because we have meters per second we have inverse seconds. And so we should recall that one in verse second, one over second is equivalent to us to the negative one. Which is also another way of writing inverse seconds, which is equivalent to a hurts just one hurts. So this will ensure we have the proper units for our frequency calculations. So beginning with our first element chromium, we'll just solve that for below Down here we're going to calculate its wavelength. So we would say that the wavelength of chromium is equal to we're going to take the speed of light which we recall is 3.0 times 10 to the eighth power meters per second. And we're dividing by our wavelength which above we converted to 2. times 10 to the negative 10th power meters. And in our calculators we should get a value for our frequency equal to 1.310 times 10. Sorry, that's times 10 To the 18th power. And when we cancel out our units we can get rid of meters leaving us with inverse seconds here, which we understood above is equivalent to one hurts. So we have units of hertz for our frequency value. So moving on to our frequency for iron, we would say that that's equal to the speed of light 3.0 times 10 to the eighth power meters per second divided by its frequency above listed as one point 937 times 10 to the negative 10th power meters which we converted equal to in our calculators, we're going to get a value of 1.59 times 10 to the 18th power. So this is times 10 to the 18th. Power hurts because we again we're left with inverse seconds as our units. So moving on to our frequency for cobalt we take the speed of light. We divide by its frequency given above as 1.790 times 10 to the negative 10th power meters. Now we cancel out meters, we get a value for the frequency equal to 1.676 times 10 to the 18th power hertz. So moving on we have our next element above which is copper. So the frequency of copper B would say is equal to the speed of light divided by its frequency above which we converted to 1.542 times 10 to the negative 10th power meters. We cancel out meters. We're left with inverse seconds and we get a frequency of 1.9 46 times 10 to the 18th power hurts. And then lastly we have our frequency of molybdenum. We take the speed of light. We divide by its frequency above which we converted to 7. times 10 to the negative 11 power meters. We canceled our units of meters were left with inverse seconds, which we understand is equivalent to 7.10 or sorry, We understand that this is going to equal in our calculators 4.2-1 times 10 to the 18th. Power hurts. So looking over all of our calculations for frequency, we just need to adjust this one here and it's actually equal to 1.549 times 10 to the 18th power hurts. So as we stated above, we are going to be plotting the square root of our frequency versus the atomic number. So recall from the periodic table that the atomic number for chromium is equal to 24 for the atomic number of iron, recalled that on the periodic table that's 26 for our atomic number of cobalt from the periodic table we get a value of 27 for our atomic number for copper. from the periodic table we have a value of 29. And for our atomic number for molybdenum we have a value of 42 from the periodic table. So now we want to go ahead and take the square root of each of our frequency calculations. And for our square root of the frequency of chromium we should get a value of 1.145 times 10 to the ninth power. So this is going to be our first y value. Moving on to the square root of the frequency of iron, we should get a value equal to 1.245 times 10 to the ninth power. Now we have the square root of our frequency of cobalt which should equal a value. And sorry about that, Which should equal a value of 1.295 times 10 to the 9th power. Next we have the square root of our frequency of copper which is going to equal a value of 1.395 times 10. And let's just make this clear, 3 95 times 10 to the ninth power. Lastly we have the square root of molybdenum which is going to give us a value of its frequency now equal to 2.55 times 10 to the ninth power. So these are all of our y values that we should or that will be generated in our graph. And now we can go ahead and plot these X and Y values into our calculators. And we will generate the following graph which I will paste below here. So here it is the graph that we should be able to generate in our calculators based on our X values being our atomic numbers and our Y values being our square root of our frequencies that we calculated above. So now we're going to go ahead. And the key thing here is our equation of our line from our graph which I will write out in a bigger format here. So we have y equal to 5.55 or sorry 0599 times 10 to the seventh power. X. And then this is going to be subtracted from our Y intercept 7.1082 times 10 to the seventh power. And this equation here of our line is what we're going to use to find the identity of our unknown element. So we want to solve for X. But right now we don't know what our Y value is here. So we need to recall that for our Y value. We can find that by taking our frequency to the one half power. Since in order to generate our Y values for our data, we took the square root of our frequencies. So we're gonna utilize this to find our value for Y. Which we can then use to find our frequency because again we said that frequency is equal to the speed of light divided by wavelength and above. According to the prompts for our unknown element. We have a wavelength value That is equal to 1. 62 extremes which we will convert to the proper units for wavelength by multiplying this by 10 to the negative 10th power. And that's going to give us 1.662 times 10 to the negative 10th power meters as our frequents or wavelength value. So all of these steps here are to find the identity of element X. So we're gonna go ahead and find its frequency. And as we stated, we would take the speed of light 3.0 times 10 to the eighth power meters per second divided by the wavelength for unknown element which the prompt tells us is 1.662 extremes which we converted two times 10 to the negative 10th power meters. So we're able to cancel out meters. We're left with inverse seconds for frequency which is what we want because we no one in verse second is equivalent to a hertz. And this is going to give us our frequency of 1. times 10 to the 18th power. And we have units of hurts now for the unknown elements. Frequency. And now that we have the frequency of our unknown element. We can use it to solve for our Y value for our equation of our line. And we would say that that is equal to our frequency which above we stated is 1.8051 times 10 to the 18th power hurts raised to the one half power. So this gives us a Y value equal to 1.3435 times 10 to the ninth power. So now that we have our Y value, we can find our identity of our unknown element X. To saw for X. So what we would have is our Y value plugged in as 1.3435 times 10 to the ninth power is equal to the rest of our equation of our line where we have our slope 5.599 times 10 to the seventh power times X. Which is what we're solving for minus our Y intercept of 7.1082 times 10 to the seventh power. And so we want to solve for X. And in doing so we would find that our X value is equal to a value of 27.9 which we can round to about 28. And so this would be our atomic number. We can say therefore our atomic number is equal to 28. And so when we refer to our periodic tables, we would see that this means that our identity of our element is going to be nickel. And so this would be our final answer as the identity of our unknown element. We have determined that it corresponds to atomic number 28, based on the wavelength that it generated and it is the atom nickel. So if you have any questions, just leave them down below. Otherwise, I will see everyone in the next practice video.