After a proton with an initial speed of 1.50 × 10⁵ m/s has increased its speed by accelerating through a potential difference of 0.100 kV,what is its final speed? (e = 1.60 × 10^-19C ,m_proton = 1.67 × 10^-27 kg)

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Quizplus · Accepted Answer

The kinetic energy gained by the proton is equal to the work done on it by the electric field, which is given by the equation $W = qV$, where $q$ is the charge of the proton (1.60 × 10^-19 C) and $V$ is the potential difference (0.100 kV = 100 V). The increase in kinetic energy ($\Delta KE$) can also be expressed as $\frac{1}{2}m(v_f^2 - v_i^2)$, where $m$ is the mass of the proton (1.67 × 10^-27 kg), $v_f$ is the final speed, and $v_i$ is the initial speed (1.50 × 10^7 m/s). Setting these two expressions for energy equal to each other and solving for $v_f$ gives the final speed of the proton.

After a Proton with an Initial Speed of 1