So, path of charged particle will be as shown below in the figure,

Clearly, charged particle will hit the screen, if $R > \omega$.

$\Rightarrow \left.\frac{m v}{B q} > \omega \text { [from, } B q v=\frac{m v^{2}}{R}\right]$

$\Rightarrow \frac{\sqrt{2 m K}}{B q} > \omega$

${[\therefore \text { momentum, } p=m v=\sqrt{2 m K}]}$

$\Rightarrow \frac{\sqrt{2 m q V}}{B q} > \omega$

$\text { or } \frac{2 m V}{B^{2} q} > \omega^{2} \text { or } \quad q < \frac{2 m V}{B^{2} \omega^{2}}$