Lead has a prominent x-ray emission line at 75.0 keV. (a) What is the minimum speed of an incident electron that could produce this emission line? (Hint: Recall the expression for relativistic kinetic energy given in Topic $26 .$ ) (b) What is the wavelength of a 75.0 -keV x-ray photon?
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here for the solution. The energy off photon is universally proportional to its wavelength. Now for the part a the relativistic kinetic energy off electron is giving us follows that que equal toe one by under route one minus V squared by C squared minus one M c square. Okay, here M is the mass off electron. We is the spirit off electron and see is the spirit off light? Now we rearrange the equation for V had very arranged the equation step by step on four we then we get equal toe after changing equal to see under Route one minus one by K E by EMC squared plus one toe the whole square. So now substitute 75.0 k e V for K and 511 k e v for m C square. Here we substituted the values in the equation. And so we get 1.468 Multiply by 10 to the power 8 m per second. Now therefore, the minimum speed off incident electro debt can electron that can produce 75.0 k e v emission line is 1.46 multiplied by 10 to the power 8 m per second. Now for the part B. The violent off photon is given as follow the Lambda equal Toe, etc. By E. Here at is the plank's constant, sees the speed off light and is the energy of a photon. Now substitute 6.63 multiplied by 10 to the power minus three J As for at you three, multiply by 10 to liberate meter per second foresee and 75.0 k e V for e. Here we substituted the values in the of education and sold for this. We get them. They called to 1.6575 multiplied by 10 to departments 11 m round off 23 significant figures. The wavelength off the photo is 1.66 multiplied by 10 to the power 11 AM And this is our final answer for the party. So this is a complete solution. Step by, Step in and deal. Please go through this. Thank you.