Question

What is the speed of an electron that has been accelerated from rest through a potential difference of 990 v ?

Answer 1

Answer:

K.E = P.E = qV { the final kinetic energy is equal to the change in the potential difference) q = 1.6 x 10^-19 V = 970V qV = 1552 x 10^-19 J K.E = 1/2mv^2. .1 m = mass of electron = 9.11 x 10^-31 kg K.E = 1552 x 10^-19 J (as K.E = = qV} v = sqrt[2(K.E)/m] {using 1} => sqrt[2(1552 x 10^-19)/9.11 x 10^-31] => sqrt[2(170.362 x 10^12)] => sqrt(340.724 x 10^12] => 18.4587 x 10^6 m/s the velocity of speed of the electron after being accelerated through a potential difference of 970V is 18.4587 x 10^6 m/s as same you solve 990 V oneself

Answer 2

Answer:

The speed of the electron is $18.6*10^{6} m/s$

Explanation:

Given that the electron is accelerated through the potential difference of 990V.

So the energy gained is P.E. which is equal to qV where q is the charge of the electron and V is the potential difference. Here if we evaluate we will obtain the energy as

$1.584*10^{-16} J$

Now we know that energy is conserved so the gained P.E. is converted into K.E.

The expression of K.E. is

$\frac{1}{2}mv^{2}$

If we equate this to the P.E. we will get that

$\frac{1}{2}mv^{2} = 1.584*10^{-16} J\\v^{2} = 2 * 1.584*10^{-16} /9.1 * 10^{-31} = 3.477 * 10^{14}\\v = \sqrt{3.477 * 10^{14}} = 18.6*10^{6} m/s$