Answer:
4.46 * 10^5 m/s
Explanation:
proton speed = 3.5 * 10 ^ 5 m/s
initial electrical potential = 600 V
second point electrical potential = 1000 V
Applying the law of conservation of energy to determine the speed at the second point
q* ( V2 - V1 ) = 1/2 M ( v2^2 - v1^2 )
making v2( speed at second point) subject of the formula( v2 ) = [tex]\sqrt{\frac{2q(v2-v1)}{m}+ v1^2 }[/tex]
V2 = 1000
V1 = 600
M = 1.67262178 * 10 ^-27 kg
q = 1.6 * 10 ^ -19 c
inserting the given values into the given equation
speed at the second point ( v2) = 4.46 * 10 ^5 m/s
note : V = POTENTIAL DIFFERENCE , v = speed
Answer:
The speed at the second point is [tex]v_{p_f} = 4.46*10^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The speed of the proton is [tex]v_p = 3.5*10^{5} \ m/s[/tex]
The electrical potential is [tex]V _1 = 600 V[/tex]
The second electrical potential is [tex]V_2 = 1000 V[/tex]
From the law of conservation of energy
The potential energy which is represented as [tex]PE = q \Delta V[/tex] is equal to the kinetic energy which is represented as
[tex]KE = \frac{1}{2} m (v_{p_f}^2 - v_p ^2 )[/tex]
So
[tex]q \Delta V = \frac{1}{2} m (v_{p_f}^2 - v_p ^2 )[/tex]
Making [tex]v_{p_f}[/tex] the subject we have
[tex]v_{p_f} = \sqrt{\frac{2q (V_2 - V_1)}{m} + v_p ^2}[/tex]
Where q is charge on the photon (electron ) which has a constant values of
[tex]q = 1.602 *10^{-19} C[/tex]
m is the mass of the photon (electron /proton) with a constant value of
[tex]m = 1.67262178 *10^{-27} \ kg[/tex]
Substituting values
[tex]v_{p_f} = \sqrt{\frac{2 (1.602 *10^{-19} * (1000 -600 ))}{1.67262178 *10^{-27}} + (3.5 *10^{5})^2}[/tex]
[tex]v_{p_f} = 4.46*10^{5} \ m/s[/tex]
