Question

The maximum energy of a deuteron coming out of a cyclotron is 20 MeV. The maximum energy of proton that can be obtained from this accelerator is:
A. 10 MeV
B. 20 MeV
C. 30 MeV
D. 40 MeV

Answer 1

In order to solve this numerical we need to know the working of cyclotron. In this process, the magnetic field provides the centripetal force then we can find the maximum energy of the deuteron and proton.

Complete step by step answer:
Given from data
$\ {E_d} = 20MeV \\\ {E_p} = ? \\\ \$
The force of magnet exerting on particle with charge q and velocity V and the centripetal force is given by
$\ F = qvB \\\ F = \dfrac{{m{v^2}}}{r} \\\ \$.
The maximum Output energy of particles is given by
$E = \dfrac{1}{2}\dfrac{{{B^2}{q^2}{r^2}}}{m}$
Similarly the maximum energy of deuteron and proton is given by
${E_p} = \dfrac{1}{2}\dfrac{{{B^2}{q^2}{r^2}}}{{{m_p}}}$…… (1)
${E_d} = \dfrac{1}{2}\dfrac{{{B^2}{q^2}{r^2}}}{{{m_d}}}$……. (2)
By taking a ratios of both the energies 2 divide 1
$\Rightarrow \dfrac{{{E_d}}}{{{E_p}}} = \dfrac{{\dfrac{1}{2}\dfrac{{{B^2}{q^2}{r^2}}}{{{m_d}}}}}{{\dfrac{1}{2}\dfrac{{{B^2}{q^2}{r^2}}}{{{m_p}}}}}$
$\Rightarrow \dfrac{{{E_d}}}{{{E_p}}} = \dfrac{{{m_p}}}{{{m_d}}}$ …….. (3)
The similar quantities get cancels each other
We know that the mass of proton ${m_p} = 1.6 \times {10^{ - 27}}kg$
Also mass of deuteron ${m_d} = 3.2 \times {10^{ - 27}}kg$
Substituting the values in equation 3
$\Rightarrow \dfrac{{20}}{{{E_p}}} = \dfrac{{1.6 \times {{10}^{ - 27}}}}{{3.2 \times {{10}^{ - 27}}}}$
$\Rightarrow {E_p} = 40MeV$

So, the correct answer is “Option D”.

Note:
A cyclotron is a device in which the particle accelerates with high energy which produces radioactive isotopes. In the stable condition, non-radioactive isotopes are put inside the cyclotron which accelerates charged particles to high energy in a both electric and magnetic field, a nuclear reaction occurs between the protons and the target particle producing the radioactive isotopes for different purposes such as medicine.