A cyclotron’s oscillator frequency is \(10MHz\). What should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is \(60 cm\). what is the kinetic energy (in \(MeV\)) of the proton beam produced by the accelerator.
\((e=1.60\times 10^{-19}C\),\(m_p=1.67\times 10^{-27}\mathit{kg},\) \(1\;MeV=1.6\times 10^{-31}J\)).
\((e=1.60\times 10^{-19}C\),\(m_p=1.67\times 10^{-27}\mathit{kg},\) \(1\;MeV=1.6\times 10^{-31}J\)).
Answer
\(7.46\mathit{MeV}\)
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Solution
The oscillator frequency should be same as proton’s cyclotron frequency.
According to the radius of the circular path of a particle we know that, \(r=\frac{mv}{qB}\) and also we know that, the larger the momentum the larger is the radius and bigger the circle described. If \(\omega \) is the angular frequency, then \(\nu =\omega r\) .
So, \(\omega =2\mathit{\pi \nu }=\frac{qB}m\) .
Using above equations we have,
\(B=\frac{2\pi m\upsilon }{q}\)
\(=\frac{2\times 3.14\times 1.67\times 10^{-27}\times 10^7}{1.6\times 10^{-19}}\)
\(=0.66T\)
Final velocity of proton is
\(v =r\times 2\pi \upsilon \)
\(=0.6\times 6.3\times 10^7\)
\(=3.78\times 10^7m/s\)
\(E=\frac 1 2mv^2\)
\(=\frac{1.67\times 10^{-27}\times (3.78\times 10^{7})^2}{2\times 1.6\times 10^{-19}}\)
\(=7.46\mathit{MeV}\)
According to the radius of the circular path of a particle we know that, \(r=\frac{mv}{qB}\) and also we know that, the larger the momentum the larger is the radius and bigger the circle described. If \(\omega \) is the angular frequency, then \(\nu =\omega r\) .
So, \(\omega =2\mathit{\pi \nu }=\frac{qB}m\) .
Using above equations we have,
\(B=\frac{2\pi m\upsilon }{q}\)
\(=\frac{2\times 3.14\times 1.67\times 10^{-27}\times 10^7}{1.6\times 10^{-19}}\)
\(=0.66T\)
Final velocity of proton is
\(v =r\times 2\pi \upsilon \)
\(=0.6\times 6.3\times 10^7\)
\(=3.78\times 10^7m/s\)
\(E=\frac 1 2mv^2\)
\(=\frac{1.67\times 10^{-27}\times (3.78\times 10^{7})^2}{2\times 1.6\times 10^{-19}}\)
\(=7.46\mathit{MeV}\)
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