Question

In a cyclotron, protons are to be accelerated. The radius of its D is 60 cm. and its oscillator frequency is 10 MHz. What will be the kinetic energy of the proton thus accelerated?
(Proton mass = 1.67 × 10⁻²⁷ kg, e = 1.60 × 10⁻¹⁹ C, eV = 1.6 × 10⁻¹⁹ J)

Answer 1

Data: R = 0.6 m, f = 10⁷ Hz, m_p = 1.67 x 10^-27 kg,

e = 1.6 x 10^-19 C, 1 eV = 1.6 x 10^-19 J

f = `"qB"/(2pi"m")_"p" and "KE" = ("q"^2"B"^2"R"^2)/(2"m"_"p")` 

∴ `"f"^2 = ("q"^2"B"^2)/(4pi^2"m"_"p"^2)`

∴ `("q"^2"B"^2)/(2"m"_"p") = 2pi^2 "f"^2"m"_"p"`

∴ KE = `2pi^2"f"^2"m"_"p""R"^2`

`= 2 xx 9.872 xx (10^7)^2(1.67 xx 10^-27)(0.6)^2`

`= 11.87 xx 10^-13 "J" = (11.87 xx 10^-13)/(1.6 xx 10^-19)`

`= 7.419 xx 10^6` eV

= 7.419 MeV