Question 132: Magnetism and Matter

Question

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 $$	imes$$ 10$$-$$19 C, mp = 1.67 $$	imes$$ 10$$-$$27 kg)

Accepted Answer

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

Answer

7.4 MeV

Answer

14.86 MeV

Answer

7.4 GeV

Answer

704 GeV

Magnetism and Matter

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, mp = 1.67 \times 10-$27 kg)

Choose the correct answer: