After a proton with an initial speed of $1.50 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }$ has increased its speed by accelerating through a potential difference of $0.100 \mathrm { kV }$, what is its final speed? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = \right.$ $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$ )

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The kinetic energy gained by the proton can be calculated using the equation $qV$, where $q$ is the charge of the proton (1.60 x $10^{-19}$ C) and $V$ is the potential difference (0.100 kV = 100 V). The increase in kinetic energy ($\Delta KE$) is equal to $1.60 \times 10^{-19} \times 100 = 1.60 \times 10^{-17}$ J. The initial kinetic energy ($KE_i$) of the proton can be calculated using $\frac{1}{2}mv^2$, where $m$ is the mass of the proton (1.67 x $10^{-27}$ kg) and $v$ is the initial speed (1.50 x $10^5$ m/s). $KE_i = \frac{1}{2} \times 1.67 \times 10^{-27} \times (1.50 \times 10^5)^2 = 1.88 \times 10^{-17}$ J. The final kinetic energy ($KE_f$) is the sum of the initial kinetic energy and the increase in kinetic energy: $KE_f = KE_i + \Delta KE = 1.88 \times 10^{-17} + 1.60 \times 10^{-17} = 3.48 \times 10^{-17}$ J. To find the final speed ($v_f$), we use $KE_f = \frac{1}{2}mv_f^2$, solving for $v_f$ gives $v_f = \sqrt{\frac{2 \times KE_f}{m}} = \sqrt{\frac{2 \times 3.48 \times 10^{-17}}{1.67 \times 10^{-27}}} = 2.04 \times 10^5$ m/s.

After a Proton with an Initial Speed Of 1.50×105 m/s1.50 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }1.50×105 m/s

After a Proton with an Initial Speed Of $1.50 \times 10 ^ { 5 } \mathrm {~m} / \mathrm { s }$