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]
