Respuesta :
Answer:
a) 714.45 Volts
b) 535.84 Volts
c) 178.125 Volts
Explanation:
a)
v₀ = initial speed of the proton = 3.7 x 10⁵ m/s
v = final speed of proton = 0 m/s
m = mass of the proton = 1.67 x 10⁻²⁷ kg
ΔV = Potential difference required
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
Using conservation of energy
q ΔV = (0.5) m (v₀² - v²)
(1.6 x 10⁻¹⁹) ΔV = (0.5) (1.67 x 10⁻²⁷) ((3.7 x 10⁵)² - (0)²)
ΔV = 714.45 Volts
b)
v₀ = initial speed of the proton = 3.7 x 10⁵ m/s
v = final speed of proton = (0.5)v₀ = (0.5) (3.7 x 10⁵) = 1.85 x 10⁵ m/s
m = mass of the proton = 1.67 x 10⁻²⁷ kg
ΔV = Potential difference required
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
Using conservation of energy
q ΔV = (0.5) m (v₀² - v²)
(1.6 x 10⁻¹⁹) ΔV = (0.5) (1.67 x 10⁻²⁷) ((3.7 x 10⁵)² - (1.85 x 10⁵ )²)
ΔV = 535.84 Volts
c)
v₀ = initial speed of the proton = 3.7 x 10⁵ m/s
m = mass of the proton = 1.67 x 10⁻²⁷ kg
K₀ = initial kinetic energy = (0.5) m v₀² = (0.5) (1.67 x 10⁻²⁷) (3.7 x 10⁵)² = 1.14 x 10⁻¹⁶ J
K = final kinetic energy = (0.5) K₀ = (0.5) (1.14 x 10⁻¹⁶) = 0.57 x 10⁻¹⁶ J
ΔV = Potential difference required
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
Using conservation of energy
q ΔV = (0.5) (K₀ - K )
(1.6 x 10⁻¹⁹) ΔV = (0.5) ((1.14 x 10⁻¹⁶) - (0.57 x 10⁻¹⁶))
ΔV = 178.125 Volts
