speed of an electron

logistic_guy

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What is the speed of an electron when it hits a television screen after being accelerated by the 25,000 V\displaystyle 25,000 \ \text{V} of the picture tube?
 
What is the speed of an electron when it hits a television screen after being accelerated by the 25,000 V\displaystyle 25,000 \ \text{V} of the picture tube?
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
Thank you Sir khan.

Let us first calculate the rest energy of the electron.

E=mc2=9.109×1031(299792458)2=8.187×1014 J\displaystyle E = mc^2 = 9.109 \times 10^{-31}(299792458)^2 = 8.187 \times 10^{-14} \ \text{J}

Let us convert it to eV\displaystyle \text{eV}.

E=8.187×10141.602×1019=511000 eV\displaystyle E = \frac{8.187 \times 10^{-14}}{1.602 \times 10^{-19}} = 511000 \ \text{eV}

Now we can calculate the electron speed.

K=mc2[11v2c21]\displaystyle K = mc^2\left[\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1\right]

25000=511000[11v2(299792458)21]\displaystyle 25000 = 511000\left[\frac{1}{\sqrt{1 - \frac{v^2}{(299792458)^2}}} - 1\right]

This gives:

v=90489723 m/s=90489723299792458c=0.302c\displaystyle v = 90489723 \ \text{m/s} = \frac{90489723}{299792458}c = 0.302c

💪:cautious:😒
 
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