A cyclotron is working at a frequency of 10 MHz. If the radius of its dees is 60 cm. The maximum kinetic energy of accelerated proton will be :

(Take : e = 1.6 $\times$ 10^$-$19 C, m_p = 1.67 $\times$ 10^$-$27 kg)

<p>Given,</p> <p>$f = 10 \times {10^6}$ Hz</p> <p>$r = 0.6$ m</p> <p>Charge on proton (q) = e</p> <p>We know,</p> <p>Radius $(r) = {{mv} \over {qB}}$</p> <p>$= {{\sqrt {2mk} } \over {eB}}$ ..... (1)</p> <p>Also, we know,</p> <p>Cyclotron oscillation frequency should be equal to the pendulum evolution frequency.</p> <p>$\therefore$ $f = {{eB} \over {2\pi m}}$</p> <p>$\Rightarrow eB = 2\pi mf$</p> <p>Putting this value of eB in equation (1) we get</p> <p>$r = {{\sqrt {2mk} } \over {2\pi mf}}$</p> <p>$\Rightarrow {r^2} = {{2mk} \over {4{\pi ^2}{m^2}{f^2}}}$</p> <p>$\Rightarrow k = 2{\pi ^2}m{f^2}{r^2}$</p> <p>$$ = 2 \times {\left( {{{22} \over 7}} \right)^2} \times 1.67 \times {10^{ - 27}} \times {\left( {10 \times {{10}^6}} \right)^2} \times {\left( {0.6} \right)^2}$$ J</p> <p>$= 1.2 \times {10^{ - 12}}$ J</p> <p>$= {{1.2 \times {{10}^{ - 12}}} \over {1.6 \times {{10}^{ - 19}}}}$ eV</p> <p>$= {{12} \over {16}} \times {10^7}$ eV</p> <p>$= 0.75 \times {10^7}$ eV</p> <p>$= 7.5 \times {10^6}$ eV</p> <p>= 7.5 MeV</p>