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Asked by Pritom Last Modified
As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's dive into the physics question you've presented.
To determine which particle is faster, we can use the kinetic energy formula:
K.E=12mv2K.E=21mv2
Where:
Given that the kinetic energies of the electron and the proton are 10 keV and 100 keV respectively, we can rearrange the kinetic energy formula to solve for velocity:
v=2K.Emv=m2K.E
Let's calculate the velocities of both particles:
For the electron: velectron=2×10×1.60×10−179.11×10−31velectron=9.11×10−312×10×1.60×10−17
For the proton: vproton=2×100×1.60×10−171.67×10−27vproton=1.67×10−272×100×1.60×10−17
Calculating these values, we get: velectron≈6.03×106 m/svelectron≈6.03×106 m/s vproton≈4.74×105 m/svproton≈4.74×105 m/s
So, the electron is faster than the proton. To find the ratio of their speeds, we divide the velocity of the electron by the velocity of the proton:
Ratio=velectronvprotonRatio=vprotonvelectron
Ratio≈6.03×1064.74×105≈12.7Ratio≈4.74×1056.03×106≈12.7
Therefore, the electron is approximately 12.7 times faster than the proton.
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