A Proton with a Speed Of $2.0 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }$

A proton with a speed of $2.0 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }$ accelerates through a potential difference and thereby increases its speed to $4.0 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }$ . Through what magnitude potential difference did the proton accelerate? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} \right)$

A) $840 \mathrm {~V}$
B) $1000 \mathrm {~V}$
C) $100 \mathrm {~V}$
D) $210 \mathrm {~V}$
E) $630 \mathrm {~V}$