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1 September, 19:49

A proton 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)

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  1. 1 September, 20:18
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    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
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