Answer
a) $\Delta V=0.84KV$
b) $\Delta v=0.63KV$
c) $\Delta V=0.42KV$
Work Step by Step
(a) We know that
$\frac{1}{2}mv_i^2+eV_i=\frac{1}{2}mv_f^2+eV_f$
$\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2=e(V_f-V_i)=e\Delta V$
This simplifies to:
$\Delta V=\frac{m(v_i^2-v_f^2)}{2e}$
We plug in the known values to obtain:
$\Delta V=\frac{(1.673\times 10^{-27})[(4.0\times 10^5)^2-(0)^2]}{2(1.6\times 10^{-19})}$
$\Delta V=0.84KV$
(b) We know that
$\Delta V=\frac{1}{2}[v_i^2-(\frac{1}{2}v_i)^2]=\frac{3mv_i^2}{8e}$
We plug in the known values to obtain:
$\Delta v=\frac{3(1.673\times 10^{-27})(4.0\times 10^5)^2}{8(1.6\times 10^{-19})}$
$\Delta v=0.63KV$
(c) We know that
$K.E_i+eV_i=\frac{1}{2}K_i+eV_f$
This simplifies to:
$\Delta V=\frac{K.E_i}{2e}$
$\implies \Delta V=\frac{mv_i^2}{2e}$
We plug in the known values to obtain:
$\Delta V=\frac{(1.673\times 10^{-27})(4.0\times 10^5)^2}{2(1.6\times 10^{-19})}=0.42KV$
