Moseley established the concept of atomic number by studying
X ra...

Question

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
(a) Calculate the frequency, n, of the X rays emitted by each
of the elements, in Hz.

Accepted Answer

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 particular elements. And we need to calculate for each element, the frequency in hertz of the X ray that is emitted. So we should recall that frequency is represented by the symbol here. However, when we look at our data were given only wavelength values and recall that wavelength is represented by the symbol lambda and is typically expressed in the units of meters. However, were given units in extremes. So we're going to recall the conversion factor that one angstrom is equivalent to 10 to the negative 10th power meters. And so we would be able to cancel out this unit of angstrom by just multiplying by 10 to the negative 10th power to each element and just adding the unit meter for the last element molybdenum. We're going to multiply by 10 to the negative 10th power in our calculators. And this is going to give us our units and meters for wavelength as zero point or rather seven point 107 times 10 to the negative 11th power. And then we have units of meters for this wavelength. So now that we have all of our units for wavelength in meters, we again need to get to frequency. So what we should do is recall that to sulfur frequency. We would take our speed of light C and divide that by our wavelength lambda. Now we should also recall that our units for the speed of light is a constant value of 3.0 times 10 to the eighth power meters per second. And as you can see, we have seconds in the denominator here. So as we stated in the prompt frequency is expressed in units of hurts. And we should recall that one hurts is equivalent to one over seconds or an inverse second. So we're going to utilize the units and speed of light to get to our units of hertz for our frequency. So first solving for the frequency of our first element chromium, we would take its wavelength in the denominator which we converted to 2.291 times 10 to the negative 10th power meters. And then above will plug in our speed of light constant 3.0 times 10 to the eighth power meters per second. So this way we'll be able to cancel out meters were left with inverse seconds. In the denominator, which we understand is equivalent to 2.291 or sorry about that equivalent to 1. 1.310 times To the 18th. Power hurts. And this is our unit for frequency. So this is for chromium. And now we're going to continue on to our second element to find the frequency of iron. So we would take again the speed of light 3.0 times 10 to the eighth power meters per second. And in our denominator we plug in the above wavelength as 1.937 times 10 to the negative 10th power meters. We cancel out meters were left with inverse seconds. And then this is going to give us our frequency equal to 1.549 times 10 to the 18th power hertz. So moving on to our frequency of cobalt, we have again our speed of light 3.0 times 10 to the eighth power meters per second. We plug in its wavelength above which we converted to 1. times 10 to the negative 10th power meters. We cancel out meters. We're left with inverse seconds in the denominator which we understand is equivalent to 1.676 times 10 to the 18th power hurts as its frequency. And now moving on to our frequency for copper, we take again the speed of light. We plug in the wavelength above which we converted for copper to 1.51 point 542 times 10 to the negative 10th power meters. We cancel out meters were left with inverse seconds and we get our frequency for copper equal to 1.9, 46 times 10 to the 18th. Power hurts. And then this leaves us with our last step, which is our frequency for molybdenum equal to the speed of light divided by its wavelength above which we determined as 7.107 times 10 to the negative 11th power meters. We cancel out meters were left with inverse seconds in the denominator, which we understand is equal to 4.2 to 1 times 10 to the 18th Power hurts for the frequency of molybdenum. So for our final answers, we're going to highlight each of our frequencies for our elements. So everything highlighted in yellow represents our final answers for the frequencies and hurts for each of our given elements according to their X rays that they admit. 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.

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