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3 years ago

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

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

6 0

Answer:

$-2.00876\times 10^{-18}\ J$

Explanation:

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

m = Mass of electron = $9.11\times 10^{-31}\ kg$

Work done would be done by

$W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J$

The work required to stop the electron is $-2.00876\times 10^{-18}\ J$

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Answer is explained in the explanation section below.

Explanation:

Solution:

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