Explanation:
We need to find the speed of a 3.1eV proton and a 3.1eV electron.
For proton, using conservation of energy such that,
[tex]\dfrac{1}{2}m_pv_p^2=eV\\\\v_p^2=\dfrac{2eV}{m_p}\\\\v_p=\sqrt{\dfrac{2eV}{m_p}}[/tex]
[tex]m_p[/tex] is mass of proton
[tex]v_p=\sqrt{\dfrac{2\times 3.1\times 1.6\times 10^{-19}\ J}{1.67\times 10^{-27}\ kg}}\\\\v_p=2.43\times 10^5\ m/s[/tex]
For electron,
[tex]\dfrac{1}{2}m_ev_e^2=eV\\\\v_e^2=\dfrac{2eV}{m_e}\\\\v_e=\sqrt{\dfrac{2eV}{m_e}}[/tex]
[tex]m_e[/tex] is mass of proton
[tex]v_e=\sqrt{\dfrac{2\times 3.1\times 1.6\times 10^{-19}\ J}{9.1\times 10^{-31}\ kg}}\\\\v_e=1.44\times 10^6\ m/s[/tex]
Hence, this is the required solution.
