Question

1. A proton has a speed of 1.5 x 109 m/s at a point where the electrical potential is 1000 V. It moves through a point where the electric potential is 900 V. What is its speed at this second point? (e = 1.60 x 10-19, mproton = 1.67 x 10-27 kg) 4.16 x 1010 m/s 2.04 x 10'0 m/s 02.04 10 m/s O 0.53 x 10ʻm/s 3.55 x 10 m/s

Answer 1

Answor!- Given that, Initial speed, t = 1.5x 105 m/s Initial potential, V = 1000x final potential, V = 900 v. charge of proton, e = 1.6081014 C -27 of proton, m= 1.67x10 Initial initial where P.E. Vi e KE = kinetic energy PE - Potential energy energy Im 11? + V, e mass 2 KE - 12 m Total be U. 2 let the speed of the proton at second point So Kinetic energy. I me? Potential energy = ve Total energy = 1 m uz tke According to conservation of energy energy Final energy 2 mV, tve Im the * *AV.243 10.22 + 1000 x -19 -27 à x 1.67 X 10 X (1.5810 t 1000 x 1.60X10 -27 -19 -x4.67810 x U t 900 x 1.60810 2

-11 1.887 10 - 11 +16810 0.635x103, 14.4810 -17 2 ² 17.88810-17 1 = 0.835 x 10 x 4 14. U X10 2 -27 7 0.835810 x Alz 2 = 3.488 10-17 3.48810-17 m Vs 0.835 X ID -27 2 4.167 x 10 5 > N2 2.048 10 m/s so the speed at second point is 2.04 x 103 mys