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How fast must an electron be moving if all its kinetic energy is lost to a single $\mathrm{x}$ -ray photon (a) at the high end of the $\mathrm{x}$ -ray electromagnetic spectrum with a wavelength of $1.00 \times 10^{-8} \mathrm{m}$ and (b) at the low end of the x-ray electromagnetic spectrum with a wavelength of $1.00 \times 10^{-13} \mathrm{m} ?$

a. 0.0220 c

b. v=0.9992 c

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{'transcript': "I'm going to let the kinetic energy of the electron 1/2 I'm so hee he squared equals the energy of a photon which would be equal to planks constant times the speed of light divided by the wavelength. We can then solve for V so V would then be equal to the square root of two times planks. Constant times the speed of light divided by the wavelength times the mass of the electron. This would be equal to we can say two times h c on divided by lambda I'm sub e C squared, multiplied by C squared, essentially multiplying This equation by C squared over C squared so the most, the speed of light squared, divided by the speed of light squared. And so this would be equal to the square root of two times planks. Constant, um, divided by Lambda Times. The mass of the electron times C squared with a factor of C. And so now we can multiply. It V would be equal to the square root rather weaken first, say the speed of light 2.998 times 10 to the eighth meters per seconds. We're going to multiply this by two have it list, right. The denominator 1st 590 nano meters multiplied by the energy M C squared, which would be 511 times 10 to the third electron volts. And then for the, uh, the numerator, we would have two times Well, 1240 electron volt. No, no meter. Let's reduce the square root here and we find that the velocity must be equal to 8.6 times 10 to the fifth meters per second. This would be our final answer for this is your final answer For the speed of the of the electron, that is the end of the solution. Thank you for watching."}

Carnegie Mellon University