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Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 106 V. m/s

User Elissa
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Final answer:

The speed of the proton after it is accelerated through an electric potential difference of 106 V is approximately 6.02 x 10^6 m/s.

Step-by-step explanation:

To calculate the speed of a proton that is accelerated from rest through an electric potential difference of 106 V, we can use the equation:

v = √(2qV/m)

Where v is the speed of the proton, q is the charge of the proton (1.6 x 10^(-19) C), V is the electric potential difference (106 V), and m is the mass of the proton (1.67 x 10^(-27) kg).

Substituting the values into the equation, we get:

v = √(2 x 1.6 x 10^(-19) C x 106 V / 1.67 x 10^(-27) kg) = 6.02 x 10^6 m/s

Therefore, the speed of the proton is approximately 6.02 x 10^6 m/s.

User Suule
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