Aproton has an initial speed of 3.7×10^5 m/s . a) what potential difference is required to bring the proton to rest? (v)
b) what potential difference is required to reduce the initial speed of the proton by a factor of 2? (v)
c) what potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2? (v)

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

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat