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Physics-

General

Easy

Question

The oscillating frequency of a cyclotron is 10 MHz. If the radius of its dees is 0.5 m, the kinetic energy of a proton, which is accelerated by the cyclotron is

10.2 MeV
2.55 MeV
20.4 MeV
5.1MeV

The correct answer is: 5.1MeV

KE of charged possible in a cyclotron,
$E subscript k end subscript equals fraction numerator q to the power of 2 end exponent B to the power of 2 end exponent r to the power of 2 end exponent over denominator 2 m end fraction$
But frequency $f equals fraction numerator q B over denominator 2 pi m end fraction$
$therefore E subscript k end subscript equals fraction numerator open parentheses 2 pi m f close parentheses to the power of 2 end exponent r to the power of 2 end exponent over denominator 2 m end fraction equals 2 pi to the power of 2 end exponent m f to the power of 2 end exponent r to the power of 2 end exponent$
Or $E subscript k end subscript equals 2 cross times open parentheses 3.14 close parentheses to the power of 2 end exponent cross times 1.67 cross times 10 to the power of negative 27 end exponent cross times open parentheses 10 cross times 10 to the power of 6 end exponent close parentheses to the power of 2 end exponent cross times open parentheses 0.5 close parentheses to the power of 2 end exponent$
$equals 8.23 cross times 10 to the power of negative 13 end exponent J$
$therefore E subscript k end subscript equals fraction numerator 8.23 cross times 10 to the power of negative 13 end exponent over denominator 1.6 cross times 10 to the power of negative 19 end exponent end fraction equals 5.1 cross times 10 to the power of 6 end exponent e V equals 5.1 blank M e V$

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