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Problem 2 Medium Difficulty

(a) What is the energy in joules of an x-ray photon with wavelength $1.00 \times 10^{-10} \mathrm{m}$ ? (b) Convert the energy to electron volts. (c) If more penetrating $\mathrm{x}$ -rays are desired, should the wavelength be increased or decreased? (d) Should the frequency be increased or decreased?

Answer

a. $1.99 \times 10^{-15} J$
b. $1.24 \times 10^{4} e V$
c. $\text { decreased }$
d. $\text { decreased }$

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Shaden K.

December 18, 2020

What is the energy in joules of an x-ray photon with wavelength 1.00 ???? 10????10 m? (b) Convert the energy to electron volts. (c) If more penetrating x-rays are desired, should the wavelength be increased or decreased? (d) Should the frequency be increa

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Marshall S.

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Video Transcript

for our question were asked to find the energy in jewels of an X ray photo on with the late wavelength of 1.0 times 10 to the minus 10 meters, which I call Lamda. So, uh, the frequency at which this X ray photon is gonna be traveling is going to be equal to the speed at which is traveling. And since it's, ah, photo on its light. So it travels at the speed of light, see, which is three times 10 to the eight meters per second, and that's gonna be divided by the wavelength. So longer wavelength means shorter frequency on a shorter wavelength. Uh, or a shorter wavelength means a higher frequency. And then the energy here is equal to place constant H multiplied by the frequency and planks. Constant A 6.63 times 10 to the minus 34 jewels time. Second, it's a common constant. You can look up if you need to. So if we plug in our expression for frequency, that means energy is equal to planes, constant times the speed of light, divided by the wavelength of the photo and plugging those values. In this expression, we find that this is equal to 1.99 times 10 to the minus 15 Jules, which we can go ahead and boxing as your solution for part a part b part B assets to convert the energy which we just found in jewels into electron volts. So that conversion between Jules and Electron volts So we're just gonna multiply our energy by that conversion is one electron volt is equal to the charge of an electron, which is 1.6 times 10 to the minus 19 cool homes. So excuse me. Not cool homes, but Jules. So if we carry out this expression, we find that this is equal to 12.4 times 10 to the three electron volts, which tended the three is kilo. So we could just rewrite this as 12.4 kilovolts. Excuse me. Kill? Uh, electron volts. K E v. I was just go ahead and rewrite that to make it look nicer. Weekend box set in as our solution for part B. Okay, so now moving on for part C, let's go ahead and start a new page to do that in part C. We are asked to figure out if a more penetrating X ray is desired, if that should have an increased wavelength or a decreased labeling. So if we increase the wavelength according to our expression that we wrote in part a where energy is equal, that H C over Lambda increasing the wavelength would decrease the energy. So therefore we need to decrease the wavelength to increase the energy. So Lamda, and then we'll ride out should be decreased. We can go ahead and let's go back to page box it in his absolution for seat. And then part D similarly asked us if we want a more penetrating X ray, should the frequency be increased or decreased? Well, if we go back here to again to part A, we said the energy is equal to the planks constant h times the frequency so increasing the frequency would increase the energy. So frequency should be increased so tight that out and we can go ahead and box set in as our solution for party

University of Kansas
Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Jared E.

University of Winnipeg

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McMaster University