## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$2.5\times 10^{4}m/s$
Proton velocity, $v_{p}=5 \times 10^{4} m/s$ For proton: By the conservation of energy: Loss in energy = Gain in energy $q\Delta V=\frac{1}{2}m_{p}v_{p}^{2}$ ...(1) Now, the same potential difference is applied for $He^{+}$. Thus, Loss in energy = Gain in energy: $q\Delta V=\frac{1}{2}m_{He^{+}}v_{He^{+}}^{2}$ ...(2) From equations (1) and (2): $v_{He^{+}}=\sqrt {\frac{m_{p}}{m_{He^{+}}}}v_{p}$ $v_{He^{+}}=\sqrt {\frac{m_{p}}{4m_{p}}}v_{p}$ $v_{He^{+}}=\frac{v_{p}}{2} =\frac{ 5 \times 10^{4}}{2} = 2.5\times 10^{4}m/s$