Question 146: 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, m
p
= 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, m p = 1.67 \times 10 -27 kg)

Choose the correct answer: