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

Question

asked Nov 23, 2021 234k views

1 Answer

SimonRH · Answer 1 · 2021-11-28T02:34:03+0000

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
$-2.00876* 10^(-18)\ J$