The particle will move in a circle
When the stopping potential is equal to the potential due to the singly charged ion at that point.
$\text{eV}_0=\frac{\text{hc}}{\lambda}-\phi$
$\Rightarrow\text{v}_0=\Big(\frac{\text{hv}}{\lambda}-\phi\Big)\frac{1}{\text{e}}$
$\Rightarrow\frac{\text{ke}}{2\text{d}}=\Big(\frac{\text{hc}}{\lambda}-\phi\Big)\frac{1}{\text{e}}$
$\Rightarrow\frac{\text{ke}}{\text{2d}}=\frac{\text{hc}}{\lambda}-\phi$
$\Rightarrow\frac{\text{hc}}{\lambda}=\frac{\text{ke}^2}{2\text{d}}+\phi=\frac{\text{ke}^2+2\text{d}\phi}{2\text{d}}$
$\Rightarrow\lambda=\frac{\text{hc 2d}}{\text{ke}^2+2\text{d}\phi}=\frac{2\text{hcd}}{\frac{1}{4\pi\omega_0\text{e}^2}2\text{d}\phi}=\frac{8\pi\omega_0\text{hcd}}{\text{e}^2+8\pi\omega_0\text{d}\phi}$
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