A cyclotron's oscillator frequency is 10 MHz and the radius of its dees is 60 cm , then the kinetic energy of the proton beam produced by the accelerator is [ mproton=1.67× 10- 27 kg ]

Question

A cyclotron's oscillator frequency is 10 MHz and the radius of its dees is 60 cm , then the kinetic energy of the proton beam produced by the accelerator is [ mproton=1.67× 10- 27 kg ] (A) 9 MeV (B) 10 MeV (C) 7 MeV (D) 11 MeV

Tardigrade Admin · Accepted Answer

KE=(q2 B2 R2/2 m) Given, f=10MHz=107 Hz R=60 cm=0.6 m f=(q B/2 π m) ∴ KE=2π 2mf2R2 =2 ×(3.14)2 × 1.67 × 10-27 × 1014 × 0.36 J ≈7MeV

Q. A cyclotron's oscillator frequency is $10 M Hz$ and the radius of its dees is $60 c m$ , then the kinetic energy of the proton beam produced by the accelerator is [ $m_{p ro t o n} = 1.67 \times 1 0^{- 27} k g$ ]

$9 M e V$

$10 M e V$

$7 M e V$

$11 M e V$

$K E = \frac{q ^{2} B ^{2} R ^{2}}{2 m}$
Given, $f = 10 M Hz = 1 0^{7} Hz$
$R = 60 c m = 0.6 m$
$f = \frac{qB}{2 πm}$
$∴ K E = 2 π^{2} m f^{2} R^{2}$
$= 2 \times (3.14)^{2} \times 1.67 \times 1 0^{- 27} \times 1 0^{14} \times 0.36 J$
$\approx 7 M e V$

Q. A cyclotron's oscillator frequency is 10MHz and the radius of its dees is 60cm , then the kinetic energy of the proton beam produced by the accelerator is [ mproton​=1.67×10−27kg ]

9MeV

10MeV

7MeV

11MeV