University Physics with Modern Physics (14th Edition)

(a) $W = 7.70 \times 10^{-14} \,\text{J}$ (b) $v = 6.80 \times 10^{6} \,\text{m/s}$
(a) The work done to separate both protons is \begin{align*} W &= \dfrac{1}{4\pi \epsilon_o} (q^2) \left[\dfrac{1}{r_a} - \dfrac{1}{r_b} \right]\\ &= (9\times 10^{9} \mathrm{~N\cdot m^2/C^2})(1.6 \times 10^{-19})^2 \left[\dfrac{1}{3 \times 10^{-15}\,\text{m}} - \dfrac{1}{2 \times 10^{-10} \,\text{m}} \right]\\ &=\boxed{ -7.70 \times 10^{-14} \,\text{J}} \end{align*} (b) The work done is equal to the change in kinetic energies of the two protons and we calculate the speed $v$ of the proton from this concept by \begin{gather*} W = 2 \Delta K \\ W = 2 (\dfrac{1}{2} mv^2) \\ W = mv^2 \\ v = \sqrt{\dfrac{W}{m}} \\ v = \sqrt{\dfrac{7.70 \times 10^{-14} \,\text{J}}{1.67 \times 10^{-27} \,\text{kg}}}\\ \boxed{v = 6.80 \times 10^{6} \,\text{m/s}} \end{gather*}