Question

please anybody solve the question​

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

option b is the right answer