Answer:
(6.67 × 10⁵) m/s
Step-by-step explanation:
What will be the proton's speed a short time later when it reaches a point where the potential is -500V.
According to the law of comservation of energy, the sum of potential and kinetic energy of the proton at any point in time is always constant.
P.E₁ + K.E₁ = P.E₂ + K.E₂
P.E₁ = U₁ = qV₁
q = charge on the proton = (1.602 × 10⁻¹⁹) C
V₁ = 1500 V
P.E₁ = U₁ = (1.602 × 10⁻¹⁹) × (1500) = (2.403 × 10⁻¹⁶) J
K.E₁ = (1/2)mv₁²
m = mass of a proton = (1.673 × 10⁻²⁷) kg
v₁ = (2.5 × 10⁵) m/s
K.E₁ = (1/2)(1.673×10⁻²⁷)(2.5 × 10⁵)² = (5.23 × 10⁻¹⁷) J
P.E₂ = U₂ = qV₂
q = charge on the proton = (1.602 × 10⁻¹⁹) C
V₂ = -500 V
P.E₂ = U₂ = (1.602 × 10⁻¹⁹) × (-500) = (-8.01 × 10⁻¹⁷) J
K.E₂ = ?
P.E₁ + K.E₁ = P.E₂ + K.E₂
(2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) = (-8.01 × 10⁻¹⁷) + K.E₂
K.E₂ = (2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) + (8.01 × 10⁻¹⁷) = (3.727 × 10⁻¹⁶) J
K.E₂ = (1/2)mv₂²
m = (1.673 × 10⁻²⁷) kg
v₂ = ?
(3.727 × 10⁻¹⁶) = (1/2)(1.673 × 10⁻²⁷)v₂²
v₂² = (4.455 × 10¹¹)
v₂ = 667,457.863839 m/s = (6.67 × 10⁵) m/s
Hope this Helps!!!!