Question
An electron is moving at $10^{6} \mathrm{~m} / \mathrm{s}$ and you wish to measure its energy to an accuracy of $\pm 0.01 \%$. What's the minimum time necessary for this measurement?
Step 1
The kinetic energy of an electron can be calculated using the formula: KE = (1/2) * m * v^2 where m is the mass of the electron (9.11 * 10^-31 kg) and v is its velocity (10^6 m/s). KE = (1/2) * (9.11 * 10^-31 kg) * (10^6 m/s)^2 KE ≈ 4.57 * 10^-19 J Now, we Show more…
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