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 (d) Use the result from part (b) to predict the X-ray wavelength emitted by iron.

Video transcript

hey everyone in this example, we're told that heavy elements amid characteristics of X rays when their electrons undergo transitions were shown below the wavelength of X rays emitted by these particular elements below. And we need to calculate the frequency of each X ray that is emitted by each of the above particular elements. And we're going to use the data to predict the wavelength of the light emitted by the atoms are kony. Um we are told and hint in the prompt which is to plot the square root of frequency. So recall that frequency 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 not given frequency were given values for wavelength and we're specifically given wavelength in units of extremes. Now recall that wavelength is represented by the symbol lambda and is typically expressed in units of meters. So we're going to cancel out these units of angstrom by recalling the conversion factor that one Angstrom is equivalent to 10 to the negative 10th power meters. Until all this means is that we have to just multiply 10 to the negative 10th power and then add the unit meters for each of our wavelengths. So we're going to do so below for the last element molybdenum. We're going to multiply by 10 to the negative 10th power and we're going to get our units for wavelength in meters equal to 7.107 times 10 to the negative power meters. So we have the wavelength of all of the listed elements. But we need to find ways length for zirconium and we don't have the element zirconium listed. So that's why according to the prompt, we need to plot a graph with our given data to find data for zirconium and its wavelength. And so what this means is that our elements here and their atomic numbers we will have to make note of. So for the atomic number for chromium and we'll actually just use the color purple From the periodic table. We're going to get a value of 24 for our atomic number of iron. We have a value of 26 for cobalt we have an atomic number of 27 for copper, we have an atomic number of 29. And for Molybdenum We have an atomic number of 42. And so for our atomic numbers Z. These will be our X values for our plot. Whereas our Y values are going to be our frequency, specifically our square root of our frequency here. So these would be our Y values. And so this means that we need to find frequency since we're not given frequency were only given wavelength. So we should recall that frequency is typically expressed in units of hertz. And so we can find frequency by taking the speed of light constancy and dividing that by our wavelength. And recall that our speed of light constant is equal to 3. times 10 to the eighth power meters per second. And because we have seconds in the denominator, we can recall the conversion factor that one in verse second which is also expressed as us to the negative one is equivalent to one hertz. And this is how we will ensure that we have our frequency values in the proper units hurts. So we're going to find those calculations below. So the frequency for our first element chromium we would say is equal to the speed of light in the numerator 3.0 times 10 to the eighth power meters per second. And in the nominator we're going to plug in its wavelength of 2.291 times 10 to the negative 10th power meters which we converted above will be able to cancel out our units of meters. Leaving us with seconds in the denominator And this is going to give us our frequency equal to 1.310 times 10 to the 18th. Power hurts now because we understand that an inverse second as we have in our conversion here is equivalent to our hurts. So now we have our first y value and now we want to find our frequency of iron. Next. So again following the same steps, the speed of light divided by its wavelength above as 1.937 times 10 to the negative 10th power meters. We cancel out meters we're left with inverse seconds which we understand is equal to 1.549 times 10 to the 18th. Power hurts moving on to our next frequency or why value? We have the frequency of cobalt. And let's just make this neader. So our frequency of cobalt we understand is equal to the speed of light. 3.0 times 10 to the eighth. Power meters per second, divided by our denominator which is the wavelength of cobalt which above we converted to 1.790 times 10 to the negative 10th. Power meters, canceling out meters were left with inverse seconds and we get our frequency and hurts equal to 1.676 times 10 to the to the positive 18th power hurts moving forward. We have our frequency of copper. Next again the numerator with the speed of light. In our denominator. A wavelength in meters which we converted above for copper as 1.542 times 10 to the negative 10th Power meters canceling out our matching units. Meters were left with inverse seconds, which we understand is equal to our frequency. For copper equal to 1.9 46 times 10 to the 18th. Power hurts. And then lastly we have our frequency of molybdenum equal to the speed of light. In the numerator 3.0 times 10 to the eighth power meters per second divided by molybdenum wavelength which we converted above to 7.107 times to the negative 11 power meters canceling out meters were left with inverse seconds Which we understand is equal to our frequency of 4.2 to 1 times 10 to the 18th power hurts. And so now to make these frequencies as our Y values, we're just going to take their square root. So for the square root of our frequency of chromium we're going to get a value equal to 1. times 10 to the 19th power. So this is our y values. So for our next Y value we take the square root of our frequency of iron which we understand is equal to 1.245 times 10 to the 19th power. Then next for the frequency of cobalt we take its square root equal to 1.295 times 10 to the 19th power and correction this should say to the ninth power for each of these Y values so far. So let's make that fix here for our next frequency for copper. We take its square root and we get the Y value of 1.3 95 times 10 to the ninth power. And then lastly we have the square root of the frequency of molybdenum equal to 2.55 times 10 to the ninth power. And so now we have all of our Y values. We also above wrote out all of our X values which were our atomic number Z for each of our elements in our data set. So now we will plug these values into our calculators to generate a graph. So I'm going to paste that graph below. So this is the graph that we should be able to generate from our calculators. And the key thing here is our equation of our line of our graph. And so to write it out bigger, we would have y equal to 5. times 10 to the seventh power X as our slope. And then this is subtracted by a Y intercept 7.1082 times 10 to the seventh power. Now according to the prompt, we need to find the wavelength emitted by zirconium. And what we should make note of here is our atomic number for zirconium where Z or its atomic number is equal to 40. So this is for zirconium. And actually we can just write it in its simple form. So we have zirconium and recall that the atomic number is written in the left hand subscript. So we can say that C. Is equal to 40. And as we stated, our X values for our atomic number, Z would fit into the X value here for our equation of our line. So we can just plug in 40 in our equation of our line. So we would have Y is equal to 5.599 times 10 to the seventh power times 40 minus 7.1082 times 10 to the seventh power. And this is going to give us our Y value equal to 1.95 to 9 times 10 to the ninth power. So now that we have our Y value, we can use this to understand how to calculate the frequency of zirconium And above we took the square root of our frequencies for each of our elements in our data set. So we can say that our frequency or rather our Y value is equal to our frequency to the one half power. Where now our frequency is going to equal Y squared. And so this means that we can take our value for Y above as 1. to 9 times 10 to the ninth power. We're going to square it. And this is going to give us our frequency equal to 3.8137 times 10 to the 18th power. And recall that frequency again is represented in units of hertz or we can say inverse seconds. So we can say that this is also equal to 3.8137 times to the 18th power, inverse seconds. And so now we can go into calculating our wavelength for zirconium. So we're going to sulfur lambda. So we would recall that lambda can be found by taking the speed of light and dividing it by our frequency. And so we would recall that our speed of light is 3.0 times 10 to the eighth power meters per second. And in our denominator we're going to plug in our frequency of zirconium which is going to be 3.8137 times 10 to the 18th power inverse seconds. We want to plug in instead of hurts so that we can cancel out the inverse seconds in the numerator for our units of the speed of light. So this allows us to cancel out the seconds with inverse seconds. And we're left with units of meters, which is what we want for wavelength. And we're going to get a final value for the wavelength of zirconium equal to 7.8663 times 10 to the negative power meters. So we would say that this is our wavelength in meters. However, just to follow the format of the prompt, we're going to give the wavelength for zirconium and angstrom. So we would say that it's equal to our wavelength in m as 7.866. And sorry, so that's six, 63 times 10 to the negative 11 power meters, which we understand is equal to in our denominator 10 to the negative 10th power meters for one angstrom. And so this would allow us to cancel out our units of meters. And so in angstrom, our wavelength for zirconium is equal to a value of 0.7866 angstrom. And so this would be our wavelength here as our final answer in angstrom for zirconium. So what's highlighted in yellow is our final answer below. If you have any questions, just leave them down below, and I will see everyone in the next practice video.