Proton inside a cyclotron

$In\quad each\quad turn\quad that\quad proton\quad make,\quad it's\\ Energy\quad \Delta E=2{ q }_{ p }V\quad \\ Where\quad V=voltage\quad applied\quad to\quad it.\\ so\quad Total\quad \uparrow es\quad in\quad \\ \Delta E=n(2{ q }_{ p }V)=n(2{ q }_{ p }Ed)\\ also\quad \quad \Delta E=\Delta KE\\ \Longrightarrow n(2{ q }_{ p }Ed)=\frac { 1 }{ 2 } { m }_{ p }{ v }_{ max }^{ 2 }\longrightarrow (1)\\ since\quad R=\frac { m{ v }_{ max } }{ { q }_{ p }B } \longrightarrow (2)\\ from\quad 1\quad \& \quad 2\\ \\ n=\frac { { q }_{ p }{ B }^{ 2 }{ R }^{ 2 } }{ 4{ m }_{ p }Ed } \\ n=9\quad Ans.\\ \\ \\ \\$ .

It will exit the Dee when its radius is R (radius of Dee) or we need

$mv/qB = R$ . now each time it crosses(equivalent of 1/2 turns) it gains an energy of $qEd$ . , thus after n crosses it gets energy of $nqEd = mv^2/2$ .

Thus solve the two equations to get n as $18.022$ . ..... now two crosses constitute one turn....hence $9.0111$ . whose box is $9$ .