Question

A CYCLOTRON,S OSCILLATOR FREQUENCY IS 10 MHz. WHAT SHOULD BE THE OPERATING MAGNETIC FIELD FOR ACCELERATING PROTONS? IF THE RADIUS OF ITS DEES IS 60cm, WHAT IS THE KINETIC ENERGY OF THE PROTON BEAM PRODUCED BY THE ACCELERATOR? EXPRESS YOUR ANSWER IN MeV.

Answer 1

sorry I don't know d answer ask someone se

Answer 2

"The kinetic energy of the proton beam produced by the accelerator is 7.2 MeV

Given:

Frequency = 10 MHz

Radius of its dees = 60 cm

Solution:

Radius of the circular path, $R\quad =\quad \frac { \sqrt { 2mk } }{ qB }$

Thus, the value of k in the above equation is $k\quad =\quad \frac { q^{ 2 }B^{ 2 }R^{ 2 } }{ 2m }$

We know that, cyclotron frequency, ${ \nu }\quad =\quad \frac { qB }{ 2\pi m }$

By applying the values of k and cyclotron frequency in the Radius of the circular path. We get,

$q^{ 2 }B^{ 2 }\quad =\quad 4\pi ^{ 2 }m^{ 2 }v^{ 2 }$

So, $k\quad =\quad \frac { 1 }{ 2m } \left( 4\pi ^{ 2 }m^{ 2 }v^{ 2 } \right) R^{ 2 }$

$k\quad =\quad 2\pi ^{ 2 }mv^{ 2 }R^{ 2 }\quad Joule$

$\because \quad 1\quad J\quad =\quad \frac { 1 }{ 1.602\quad \times \quad 10^{ -19 } } \quad =\quad \frac { 1 }{ e } \quad eV$

$k\quad =\quad \left( \frac { 2\pi ^{ 2 }mv^{ 2 }R^{ 2 } }{ e } \right) eV$

By applying the values in the above equation we get,

$\Rightarrow \quad \left\{ \frac { 2\quad \times \quad 10\quad \times \quad \left( 1.67\quad \times \quad 10^{ 27 } \right) \quad \times \quad \left( 10\quad \times \quad 10^{ 6 } \right) ^{ 2 }\quad \times \quad (0.6)^{ 2 } }{ 1.6\quad \times \quad 10^{ -19}}\right\} eV$

By simplifying we get,

$k\quad =\quad 7.2\quad \times \quad 10^{ 6 }\quad eV$

$k\quad =\quad 7.2\quad MeV$"