X rays with a wavelength of 1.54 * 10-10 m are produced when a copper metal target is bombarded with high-energy electrons that have been accelerated by a voltage difference of 30,000 V. The kinetic energy of the electrons equals the product of the voltage difference and the electronic charge in coulombs, where 1 volt-coulomb = 1 J. (a) What is the kinetic energy in joules and the de Broglie wavelength in meters of an electron that has been accel-erated by a voltage difference of 30,000 V?

Video transcript

welcome back everyone. We're told that when high energy electrons that have been accelerated by the voltage difference of 25,000 volts bombard a metal sample target X rays with a wavelength of 1.74 times 10 to the negative 10 power meters are created. The product of the voltage difference and the electronic charge expressed in columns yields the kinetic energy of the electrons where one volt column equals one jewel. What are the electron's kinetic energy and jewels? And the broccoli wavelength in meters when accelerated by 25,000 voltage difference. So we want to recall the following formulas first for the kinetic energy of an electron, which according to our prompt, Can be found by taking the product of our voltage difference given as 25,000V multiplied by the electronic charge of an electron. And we're going to recall that the electronic charge of an electron is equal to 1.6 oh two times 10 to the negative 19 power columns. So we will multiply by 1.6 oh two times 10 to the negative power columns. And because we have units of volts being multiplied by columns, we're going to recall that a vault times a column is equivalent to one Juul. And so for our results here, the kinetic energy of our electron according to our prompt should equal the value of the product, which results in four point oh oh five times 10 to the negative 15 power jewels. Now, without that information being given to us in the prompt, we also want to recall our formal formula to calculate the kinetic energy of an electron where we would take one half times the mass of our electron times its velocity squared. And with this formula we recall. We also want to recall that because we need to give our wavelength for this electron according to our prompt. As our second answer, we need to recall our formula where wavelength represented by lambda is calculated by taking plank's constant H. Divided by our massive our electron multiplied by its velocity V. Meaning that we need to isolate our formula for the kinetic energy of an electron for velocity so that it's equal to the square root of two, multiplied by our kinetic energy of our electron divided by our mass of our electron. M and so everything here would be under the square root just to be clear. So let's find this velocity. What we're going to have is our square root of two multiplied by the kinetic energy of our electron, which we just determined to be 4.005 times 10 to the negative 15 power jewels. And then we're going to divide this by the mass of our electron, which Is not given to us in our prompt. But we should recall from our textbooks that it's equal to a value of 9.109 times 10 to the negative 31st power units of kg. So before we continue our calculation, we next want to recall that one jewel is equivalent. And let's make this clear. So one jewel is equivalent to one kg times meter squared divided by seconds squared. So we're going to interpret our units here as follows. And what we're going to result in when we plug this carefully into our calculators as a result of 9.3774 times 10 to the seventh power where we'll interpret our units as now meters per second because we will have interpreted our jewels as now kilograms divided by meter. Sorry, kilograms multiplied by meters squared, divided by seconds squared, meaning we would cancel out one of our units of meters and seconds and also canceling out kilograms. And so that is why we have units of meters per second here. So now we need to get to our second answer which is going to be our wavelength by incorporating this velocity into our formula for wavelength so that we can say our wavelength is equal to Planck's constant in the numerator as 6.626 times 10 to the negative 34th power jewels. Time seconds. However, we want to incorporate or interpret these units of jewels. Time seconds as now kilograms times meters squared divided by seconds, which we then want to divide by our massive our electron, which again from our textbooks is a value of 9.109 times 10 to the negative 31st power units of kilograms multiplied by our velocity of our electron, which we just calculated as 9.3774 times 10 to the seventh power units of meters per second. So canceling out our units, we see we can get rid of one of our units of meters here with meters in the denominator. We can cancel out seconds and we can also cancel out kilograms. And this leaves us with meters as our final unit for wavelength, which is what we want, which is going to result in a value of 7.757 times 10 to the negative 12 power meters. So for our final answers, we have the wave or the d broadly wavelength of our electron as 7.757 times 10 to the negative 12 power meters and our first answer for our kinetic energy of our electron equal to 4.5 times 10 to the negative 15 power jewels. So it's highlighted in yellow are two final answers to complete this example. I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video