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How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10

User MaxKlaxx
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1 Answer

3 votes

Answer:


-2.00876* 10^(-18)\ J

Step-by-step explanation:

v = Speed of electron =
2.1* 10^6\ m/s (generally the order of magnitude is 6)

m = Mass of electron =
9.11* 10^(-31)\ kg

Work done would be done by


W=K_i-K_f\\\Rightarrow W=0-(1)/(2)mv^2\\\Rightarrow W=0-(1)/(2)* 9.11* 10^(-31)* (2.1* 10^6)^2\\\Rightarrow W=-2.00876* 10^(-18)\ J

The work required to stop the electron is
-2.00876* 10^(-18)\ J

User SimonRH
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7.1k points