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A Proton That Is Initially at Rest Is Accelerated Through 500 V500 \mathrm {~V}

Question 93

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A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V500 \mathrm {~V} . What speed does the proton gain? (e=1.60×1019C,m\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m \right. proton =1.67×1027 kg) \left. = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} \right)


A) 1.1×105 m/s1.1 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }
B) 3.1×105 m/s3.1 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }
C) 2.2×105 m/s2.2 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }
D) 9.6×105 m/s9.6 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }

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