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A proton with an initial speed of 600,000 m/s is brought to rest by an electric field.1. What was the potential difference that stopped the proton?
2. What was the initial kinetic energy of the proton, in electron volts?

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Answer:

(1) the potential difference that stopped the proton is 1878.75 V

(2) the initial kinetic energy of the proton is 1878.75 eV

Explanation:

Given;

initial speed of the proton, v = 600,000 m/s

mass of proton, m = 1.67 x 10⁻²⁷ kg

(1) The work done in bringing the proton to rest is given as;

[tex]W = eV[/tex]

Apply work energy theorem;

[tex]K.E =W\\\\ \frac{1}{2} mv^2 = eV\\\\V = \frac{mv^2}{2e}[/tex]

where;

V is the potential difference

[tex]V = \frac{1.67\times 10^{-27} \times\ 600,000^2}{2 \ \times \ 1.6 \times 10^{-19}} \\\\V = 1878.75 \ V[/tex]

(2) the initial kinetic energy of the proton, in electron volts;

[tex]K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times \ 1.67\times 10^{-27} \times 600,000^2 = 3.006 \times 10^{-16} \ J\\\\K.E = \frac{3.006 \times 10^{-16} \ J \ \ eV}{1.6 \times 10^{-19} \ J} = 1878.75 \ eV[/tex]

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