Cyclotron Questions in English

(B) The angular frequency $\omega$ of a charged particle moving in a magnetic field $B$ is given by the formula $\omega = \frac{qB}{m}$.
Here,$q$ is the charge of the particle,$B$ is the magnetic field strength,and $m$ is the mass of the particle.
From this expression,it is clear that the angular frequency $\omega$ depends only on the charge,the mass of the particle,and the magnetic field.
It does not depend on the velocity or speed $v$ of the particle.
Therefore,the angular frequency is independent of the speed of the charged particle.

(C) The kinetic energy is given by $K = \frac{1}{2}mv^2$.
In a cyclotron,the radius of the circular path of a charged particle is $r_0 = \frac{mv}{qB}$.
Rearranging this for velocity,we get $v = \frac{qB r_0}{m}$.
Substituting this value of $v$ into the kinetic energy formula:
$K_{\max} = \frac{1}{2}m \left( \frac{qB r_0}{m} \right)^2$
$K_{\max} = \frac{1}{2}m \left( \frac{q^2 B^2 r_0^2}{m^2} \right)$
$K_{\max} = \frac{q^2 B^2 r_0^2}{2m}$.

(C) cyclotron is a device used to accelerate positively charged particles such as protons,deuterons,and alpha particles.
It works on the principle that a positively charged particle can be accelerated to a sufficiently high energy by repeatedly crossing an oscillating electric field,while being guided in a circular path by a strong magnetic field.
Electrons are not accelerated in a cyclotron because their mass is very small,causing them to reach relativistic speeds quickly,which leads to a loss of synchrony with the oscillating electric field.

(D) The cyclotron frequency $v$ is given by the formula $v = \frac{qB}{2\pi m}$.
Given: $B = 1\, T$,$q = 1.6 \times 10^{-19}\, C$,and $m = 9.1 \times 10^{-31}\, kg$.
Substituting the values:
$v = \frac{1 \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}$
$v = \frac{1.6}{57.148} \times 10^{12}\, Hz$
$v \approx 0.02799 \times 10^{12}\, Hz = 27.99 \times 10^9\, Hz$
$v \approx 28\, GHz$.

(C) The cyclotron frequency $f$ is given by $f = \frac{eB}{2\pi m}$.
Rearranging for the magnetic field $B$,we get $B = \frac{2\pi mf}{e}$.
The maximum velocity $v$ of the proton at the exit of the dee (radius $R$) is given by $v = \omega R = (2\pi f)R$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.
Substituting $v = 2\pi fR$ into the kinetic energy formula:
$K = \frac{1}{2}m(2\pi fR)^2 = \frac{1}{2}m(4\pi^2f^2R^2) = 2m\pi^2f^2R^2$.
Thus,$B = \frac{2\pi mf}{e}$ and $K = 2m\pi^2f^2R^2$.

(A) In a cyclotron,the charged particle is accelerated by the oscillating electric field present in the gap between the two dees.
$(i)$ The particle speeds up between the dees due to the electric field.
$(ii)$ Inside the dees,the magnetic field acts on the particle,causing it to move in a circular path with a constant speed,but its direction changes continuously.
Therefore,the particle undergoes acceleration due to the electric field in the gap.

(C) The maximum kinetic energy $KE_{\max}$ of a particle in a cyclotron is given by the formula $KE_{\max} = \frac{q^2 B^2 R^2}{2m}$, where $q$ is the charge, $B$ is the magnetic field, $R$ is the radius of the dees, and $m$ is the mass of the particle.
Since $B$ and $R$ are constant for a given cyclotron, $KE_{\max} \propto \frac{q^2}{m}$.
For a proton $(p)$: $q = e, m = m_p \implies \frac{q^2}{m} = \frac{e^2}{m_p}$.
For a deuteron $(d)$: $q = e, m = 2m_p \implies \frac{q^2}{m} = \frac{e^2}{2m_p}$.
For an alpha particle $(\alpha)$: $q = 2e, m = 4m_p \implies \frac{q^2}{m} = \frac{4e^2}{4m_p} = \frac{e^2}{m_p}$.
Comparing the values, the ratio $\frac{q^2}{m}$ is minimum for the deuteron. Therefore, the minimum kinetic energy gained is for the deuteron.

(D) The kinetic energy $(E_k)$ of a charged particle in a cyclotron is given by the formula: $E_k = \frac{q^2 B^2 r^2}{2m}$.
The frequency of the cyclotron is given by: $f = \frac{qB}{2 \pi m}$,which implies $qB = 2 \pi m f$.
Substituting $qB$ into the kinetic energy formula:
$E_k = \frac{(2 \pi m f)^2 r^2}{2m} = 2 \pi^2 m f^2 r^2$.
Given values:
$m = 1.67 \times 10^{-27} \, kg$ (mass of proton)
$f = 10 \times 10^6 \, Hz$
$r = 0.5 \, m$
Calculating $E_k$ in Joules:
$E_k = 2 \times (3.14)^2 \times (1.67 \times 10^{-27}) \times (10^7)^2 \times (0.5)^2$
$E_k = 2 \times 9.8596 \times 1.67 \times 10^{-27} \times 10^{14} \times 0.25$
$E_k \approx 8.23 \times 10^{-13} \, J$.
Converting to $MeV$:
$E_k (in \, eV) = \frac{8.23 \times 10^{-13}}{1.6 \times 10^{-19}} \approx 5.14 \times 10^6 \, eV = 5.14 \, MeV$.
Rounding to the nearest option,we get $5.1 \, MeV$.

(D) In a cyclotron,the maximum velocity $v$ of a particle is related to the radius $r$ and frequency $f$ by the formula $v = r \omega = r(2 \pi f)$.
Given:
Frequency $f = 10\, MHz = 10 \times 10^6\, Hz = 10^7\, Hz$.
Radius $r = 50\, cm = 0.5\, m$.
Substituting these values into the formula:
$v = 2 \times \pi \times r \times f$
$v = 2 \times 3.14 \times 0.5 \times 10^7$
$v = 1 \times 3.14 \times 10^7\, m/s$
$v = 3.14 \times 10^7\, m/s$.

(C) cyclotron is a particle accelerator used to accelerate charged particles,such as positive ions,to high energies. Thus,the Assertion is correct.
The cyclotron frequency $f$ is given by the formula:
$f = \frac{Bq}{2 \pi m}$
where $B$ is the magnetic field,$q$ is the charge,and $m$ is the mass of the particle.
From this formula,it is clear that the cyclotron frequency depends only on the magnetic field,the charge,and the mass of the particle. It is independent of the velocity $v$ and the radius $r$ of the orbit.
Therefore,the Reason is incorrect.

(C) cyclotron operates on the principle that the frequency of the oscillating electric field matches the cyclotron frequency of the particle,which is given by $f = \frac{qB}{2\pi m}$.
Because the mass of an electron $(m)$ is very small,it gains speed very rapidly when accelerated.
According to the theory of relativity,as the speed of the electron approaches the speed of light,its relativistic mass increases significantly $(m = \frac{m_0}{\sqrt{1 - v^2/c^2}})$.
This increase in mass causes the cyclotron frequency to change,leading to a mismatch between the frequency of the $a.c.$ source and the particle's rotation frequency in the Dees.
Consequently,the electron falls out of phase with the electric field and is no longer accelerated effectively. Thus,both the assertion and the reason are correct,and the reason is the correct explanation for the assertion.

(A) The oscillator frequency $f$ must be equal to the cyclotron frequency of the proton.
$f = \frac{qB}{2\pi m_p} \implies B = \frac{2\pi m_p f}{q}$
Substituting the values: $B = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 10^7}{1.60 \times 10^{-19}} \approx 0.656 \; T \approx 0.66 \; T$.
The maximum velocity $v$ of the proton at the exit radius $r = 0.6 \; m$ is given by $v = r\omega = r(2\pi f)$.
$v = 0.6 \times 2 \times 3.14 \times 10^7 = 3.77 \times 10^7 \; m/s$.
The kinetic energy $K$ is given by $K = \frac{1}{2} m_p v^2$.
$K = \frac{1}{2} \times 1.67 \times 10^{-27} \times (3.77 \times 10^7)^2 = 1.186 \times 10^{-12} \; J$.
Converting to $MeV$: $K = \frac{1.186 \times 10^{-12}}{1.6 \times 10^{-13}} \approx 7.41 \; MeV$.

(N/A) When a charged particle having charge $q$ and velocity $\vec{v}$ enters a uniform magnetic field $\vec{B}$,it experiences a magnetic Lorentz force given by:
$\vec{F} = q(\vec{v} \times \vec{B})$
Since the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the force $\vec{F}$ acts perpendicular to both $\vec{v}$ and $\vec{B}$. This force acts as a centripetal force,causing the particle to move in a uniform circular path.
For uniform circular motion,the magnetic force provides the necessary centripetal force:
$\text{Centripetal force} = \text{Magnetic force}$
$\frac{mv^2}{r} = qvB$
Solving for the radius $r$:
$r = \frac{mv}{qB} \quad \dots (1)$
Since linear momentum $p = mv$,we can write:
$r = \frac{p}{qB}$
This indicates that as the momentum increases,the radius of the circular path also increases.
Using the relation between linear velocity and angular velocity,$v = \omega r$:
$\frac{v}{\omega} = \frac{mv}{qB}$
$\omega = \frac{qB}{m}$
where $\omega$ is the angular frequency.
The frequency of revolution $f$ is given by:
$f = \frac{\omega}{2\pi} = \frac{qB}{2\pi m}$
This is known as the cyclotron frequency,which is independent of the speed of the particle and the radius of the orbit. This principle is used in a cyclotron to accelerate charged particles.

(N/A) cyclotron is a device used to accelerate charged particles like protons,deuterons,$\alpha$-particles,etc.,to very high energies. It was invented by $E.O. Lawrence$ and $M.S. Livingston$ in $1934$.
The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles. As the fields are perpendicular to each other,they are called crossed fields.
The cyclotron operates on the principle that the frequency of revolution of a charged particle in a magnetic field is independent of its energy.
Principle: $A$ charged particle can be accelerated to very high energies by making it pass through a moderate electric field multiple times. This is achieved with the help of a perpendicular magnetic field,which forces the charged particle into a circular motion. The frequency of this circular motion (cyclotron frequency) is independent of the speed of the particle and the radius of the circular orbit.

(N/A) The construction of a cyclotron is as shown in the diagram.
It consists of two small hollow,metallic half-cylinders,$D_{1}$ and $D_{2}$,called 'dees' because they are in the shape of the letter $D$.
They are mounted inside a vacuum chamber between the poles of a powerful electromagnet.
The dees are connected to a source of high-frequency alternating voltage of a few hundred kilovolts.
The beam of charged particles to be accelerated is injected into the dees near their center $P$,in a plane perpendicular to the magnetic field.
The charged particles are pulled out of the dees by a negatively charged deflecting plate through a window $W$.
The whole device is kept in a high vacuum so that air molecules do not collide with the charged particles.

(N/A) The figure shows a schematic view of the cyclotron.
Inside the metal boxes (dees),the particle is shielded and is not acted upon by the electric field. The electric field prevails only in the gap between the two dees.
The magnetic field acts on the particle and forces it to move in a circular path inside the dee.
Every time the particle moves from one dee to another,it is accelerated by the electric field.
The polarity of the electric field is reversed in synchronization with the circular motion of the particle.
This ensures that the particle is always accelerated by the electric field. Each acceleration increases the kinetic energy of the particle. As energy increases,the radius of the circular path increases,resulting in a spiral path.
Charged particles (e.g.,protons) move in a semicircular path in one of the dees and arrive in the gap between the dees in a time interval of $\frac{T}{2}$,where $T$ is the time period of revolution.
$T = \frac{1}{v_{c}} = \frac{2 \pi m}{q B}$ or $v_{c} = \frac{q B}{2 \pi m} \quad \dots (1)$
This frequency is called the cyclotron frequency and is denoted by $v_{c}$. It is independent of the speed,momentum,and kinetic energy of the particle.
The uses of a cyclotron are as follows:
$(1)$ The high-energy particles produced in a cyclotron are used to bombard nuclei to study nuclear reactions and investigate nuclear structure.
$(2)$ It is used to implant ions into solids to modify their properties or to synthesize new materials.
$(3)$ It is used to produce radioactive isotopes,which are utilized in hospitals for medical diagnosis and treatment.

(A) cyclotron is a type of particle accelerator.
It uses a magnetic field to keep charged particles in a circular path and an oscillating electric field to increase their kinetic energy (accelerate them) each time they cross the gap between the two dees.
Therefore,the primary purpose of a cyclotron is to accelerate charged particles like protons,deuterons,and alpha particles to high energies.

(N/A) In a cyclotron,'dees' are two hollow,semi-circular metal chambers shaped like the letter '$D$'.
They are placed in a vacuum chamber with a small gap between them.
$A$ high-frequency alternating voltage is applied across these dees.
The interior of the dees is a field-free region,meaning the electric field inside them is zero.
As a charged particle enters a dee,it moves in a semi-circular path due to the presence of a uniform perpendicular magnetic field.
When the particle reaches the gap,the electric field accelerates it,increasing its kinetic energy before it enters the next dee.

(N/A) The resonance condition for a cyclotron occurs when the frequency of the applied alternating electric field $(f_a)$ matches the cyclotron frequency $(f_c)$ of the charged particle moving in a magnetic field.
The cyclotron frequency is given by the formula:
$f_c = \frac{qB}{2\pi m}$
Where:
$q$ = charge of the particle
$B$ = magnetic field strength
$m$ = mass of the particle
Therefore,the resonance condition is:
$f_a = f_c = \frac{qB}{2\pi m}$

(A) When a charged particle enters a perpendicular magnetic field,the magnetic force provides the necessary centripetal force for circular motion.
$F_m = F_c$
$qvB = \frac{mv^2}{R}$
Rearranging for angular velocity $\omega = \frac{v}{R}$:
$\frac{v}{R} = \frac{qB}{m}$
Since $\omega = \frac{v}{R}$ and $q = e$,we have $\omega = \frac{eB}{m}$.
Now,checking the dimensions:
$[\omega] = \frac{[e][B]}{[m]}$
From $F = qvB$,the dimension of $B$ is $[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{IT \cdot LT^{-1}} = [M I^{-1} T^{-2}]$.
Substituting the dimensions:
$[\omega] = \frac{[I T] \cdot [M I^{-1} T^{-2}]}{[M]} = \frac{M I^0 T^{-1}}{M} = [T^{-1}]$.
Thus,the dimensions of cyclotron frequency are $[T]^{-1}$.

(N/A) The frequency of the radio frequency $(rf)$ field is given by $f = \frac{qB}{2\pi m}$. The time period of the particle in the cyclotron is $T = \frac{2\pi m}{qB}$.
If the frequency of the $(rf)$ field is doubled,the oscillator frequency becomes $2f$,which means the time period of the $(rf)$ field becomes $T' = \frac{T}{2}$.
Since the time taken by the particle to complete a semi-circle inside a dee is $t = \frac{T}{2}$,the particle will now reach the gap between the dees before the electric field has completed its cycle.
Consequently,the particle will not be accelerated consistently at the gap,as the electric field polarity will not match the arrival of the particle. This disrupts the resonance condition required for the cyclotron to function,and the particle will not gain energy effectively.

(C) The kinetic energy gained by a proton after $n$ revolutions in a cyclotron is given by $K = n \times (2qV)$,where $V$ is the maximum accelerating potential per gap and there are two gaps per revolution.
Given:
$V = 12 \times 10^3 \, V$
$q = 1.6 \times 10^{-19} \, C$
$m_p = 1.67 \times 10^{-27} \, kg$
$v = \frac{c}{6} = \frac{3 \times 10^8}{6} = 0.5 \times 10^8 \, m/s$
Equating the kinetic energy to the work done:
$n(2qV) = \frac{1}{2} m_p v^2$
$n = \frac{m_p v^2}{4qV}$
$n = \frac{1.67 \times 10^{-27} \times (0.5 \times 10^8)^2}{4 \times 1.6 \times 10^{-19} \times 12 \times 10^3}$
$n = \frac{1.67 \times 10^{-27} \times 0.25 \times 10^{16}}{76.8 \times 10^{-16}}$
$n = \frac{0.4175 \times 10^{-11}}{76.8 \times 10^{-16}} \approx 543.6$
Rounding to the nearest integer,the number of revolutions is $543$.

(C) The radius $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula $r = \frac{mv}{qB}$.
Since the kinetic energy $K$ is related to momentum $p$ by $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.
Substituting $p = mv$ into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.
This shows that the radius is directly proportional to the square root of the kinetic energy: $r \propto \sqrt{K}$.
Given that the new kinetic energy $K_n = 4K_0$,where $K_0$ is the initial kinetic energy.
The ratio of the new radius $r_n$ to the original radius $r_0$ is $\frac{r_n}{r_0} = \sqrt{\frac{K_n}{K_0}} = \sqrt{\frac{4K_0}{K_0}} = \sqrt{4} = 2$.
Therefore,the ratio is $2: 1$.

(B) The kinetic energy $K$ of a particle of charge $q$ and mass $m$ in a cyclotron with magnetic field $B$ and radius $r$ is given by the formula:
$K = \frac{q^2 B^2 r^2}{2m}$
Given:
$q = 1.6 \times 10^{-19}\,C$
$B = 1.0\,T$
$r = 60\,cm = 0.6\,m$
$m = 1.6 \times 10^{-27}\,kg$
Substituting the values:
$K = \frac{(1.6 \times 10^{-19})^2 \times (1.0)^2 \times (0.6)^2}{2 \times 1.6 \times 10^{-27}}$
$K = \frac{2.56 \times 10^{-38} \times 0.36}{3.2 \times 10^{-27}}$
$K = 0.8 \times 10^{-11} \times 0.36 = 0.288 \times 10^{-11}\,J$
To convert Joules to $MeV$,divide by $1.6 \times 10^{-13}$:
$K = \frac{0.288 \times 10^{-11}}{1.6 \times 10^{-13}} = 0.18 \times 10^2 = 18\,MeV$

(D) The cyclotron frequency $f$ is given by the formula: $f = \frac{qB}{2 \pi m}$.
Rearranging the formula to solve for the magnetic field $B$,we get: $B = \frac{2 \pi m f}{q}$.
Given values are: $f = 10 \,MHz = 10^7 \,Hz$,$m = 1.67 \times 10^{-27} \,kg$,and $q = 1.6 \times 10^{-19} \,C$.
Substituting these values into the equation:
$B = \frac{2 \times 3.14159 \times 1.67 \times 10^{-27} \times 10^7}{1.6 \times 10^{-19}}$.
$B = \frac{10.493 \times 10^{-20}}{1.6 \times 10^{-19}}$.
$B \approx 0.656 \,T$.
Thus,the correct option is $D$.

$(A)$ In a cyclotron, a particle gains energy equal to $qV$ every time it crosses the gap between the two "dees".
Since there are two gaps per revolution, the energy gained per revolution is $\Delta E = 2qV$.
Given: Potential difference $V = 100 \,kV = 10^5 \,V$.
Energy gained per revolution $= 2 \times e \times 10^5 \,V = 2 \times 10^5 \,eV$.
Target kinetic energy $E = 20 \,MeV = 20 \times 10^6 \,eV = 2 \times 10^7 \,eV$.
Number of revolutions $n = \frac{E}{\Delta E} = \frac{2 \times 10^7 \,eV}{2 \times 10^5 \,eV} = 100$.

(C) For a particle of mass $m$ and charge $q$ moving in a magnetic field $B$, the cyclotron frequency is given by $\omega = \frac{qB}{m}$.
According to Bohr's quantization condition, the angular momentum $L$ in the $n^{th}$ level is $L = n \frac{h}{2 \pi}$.
Since $L = mvr = mr^2 \omega$, we have $mr^2 \omega = n \frac{h}{2 \pi}$.
Substituting $\omega = \frac{qB}{m}$, we get $mr^2 (\frac{qB}{m}) = n \frac{h}{2 \pi}$, which simplifies to $r^2 = \frac{nh}{2 \pi qB}$.
The kinetic energy $K$ is given by $K = \frac{1}{2} mv^2 = \frac{1}{2} m(r \omega)^2 = \frac{1}{2} m r^2 \omega^2$.
Substituting $r^2$ and $\omega$, we get $K = \frac{1}{2} m (\frac{nh}{2 \pi qB}) (\frac{qB}{m})^2 = \frac{n h q B}{4 \pi m}$.
For the second level, $n = 2$, so $K = \frac{2 h q B}{4 \pi m} = \frac{qBh}{2 \pi m}$.

(B) cyclotron is a particle accelerator that uses a combination of magnetic and electric fields to accelerate charged particles. It is primarily designed to accelerate positively charged particles,such as protons,deuterons,and alpha particles,to high energies. Neutrons cannot be accelerated by a cyclotron because they lack an electric charge.

(C) The time period $T$ for a full circular orbit of a charged particle in a magnetic field $B$ is given by $T = \frac{2 \pi m}{q B}$.
Since the ion describes a semicircular path in a dee,the time taken is $t = \frac{T}{2} = \frac{\pi m}{q B}$.
From this expression,it is clear that the time taken is independent of the speed of the ion $(v)$ and the radius of the circular path $(r)$.

(A) The cyclotron frequency is given by $n = \frac{q B}{2 \pi m}$.
Rearranging for the magnetic field $B$,we get $B = \frac{2 \pi n m}{q}$.
For a proton moving in a circular path of radius $r$ inside the dees,the magnetic force provides the centripetal force: $\frac{m v^2}{r} = q v B$.
This simplifies to $m v^2 = q v B r$.
The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m v^2$.
Substituting $m v^2 = q v B r$ into the kinetic energy formula,we get $K.E. = \frac{q v B r}{2}$.

(A) The resonance condition for a cyclotron is that the frequency of the oscillator '$v$' must be equal to the cyclotron frequency: $v = \frac{eB}{2 \pi m}$.
From this,we can express the magnetic field as $B = \frac{2 \pi m v}{e}$.
The radius of the path of the proton in the cyclotron is given by $R = \frac{mv}{eB}$,where '$v_{p}$' is the velocity of the proton.
Rearranging for velocity: $v_{p} = \frac{eBR}{m}$.
Substituting the expression for '$B$': $v_{p} = \frac{e}{m} \times \left( \frac{2 \pi m v}{e} \right) \times R = 2 \pi v R$.
The kinetic energy $(K.E.)$ of the proton is given by $K.E. = \frac{1}{2} m v_{p}^{2}$.
Substituting '$v_{p}$': $K.E. = \frac{1}{2} m (2 \pi v R)^{2} = \frac{1}{2} m (4 \pi^{2} v^{2} R^{2}) = 2 m \pi^{2} v^{2} R^{2}$.

(D) The maximum kinetic energy $E_K$ gained by a charged particle in a cyclotron is given by the formula:
$E_K = \frac{q^2 B^2 R^2}{2m}$
where,
$q$ is the charge of the particle,
$B$ is the magnetic field intensity,
$R$ is the maximum radius of the orbit (radius of the dees),
and $m$ is the mass of the particle.
From the formula,it is evident that $E_K$ depends on $q$,$B$,$R$,and $m$.
However,$E_K$ does not depend on the frequency of revolution $(f = \frac{qB}{2\pi m})$,as the frequency is determined by the cyclotron resonance condition and is independent of the energy or radius of the particle's path.

(A) In a cyclotron,the frequency of the alternating electric field $f$ is equal to the cyclotron frequency $f_c = \frac{eB}{2\pi m}$.
From this,the magnetic field $B$ is given by $B = \frac{2\pi mf}{e}$.
The maximum velocity $v$ of the proton at the exit radius $R$ is given by $v = \omega R = (2\pi f)R$.
The kinetic energy $K.E.$ is given by $K.E. = \frac{1}{2}mv^2$.
Substituting $v = 2\pi f R$,we get $K.E. = \frac{1}{2}m(2\pi f R)^2 = \frac{1}{2}m(4\pi^2 f^2 R^2) = 2\pi^2 mf^2 R^2$.
Thus,the magnetic field is $\frac{2\pi mf}{e}$ and the kinetic energy is $2\pi^2 mf^2 R^2$.

(B) cyclotron is a particle accelerator that uses a magnetic field to keep charged particles in a circular path and an alternating electric field to increase their kinetic energy.
It is primarily used to accelerate positively charged particles such as protons,deuterons,and alpha particles.
Neutrons cannot be accelerated by a cyclotron because they are electrically neutral and do not experience the Lorentz force $(F = q(v \times B))$ required for circular motion or the electric force $(F = qE)$ required for acceleration.

(A) In a cyclotron,the charged particle moves in a circular path inside the dees ($D_1$ and $D_2$) due to a perpendicular magnetic field. The magnetic force acting on the particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$,which is always perpendicular to the velocity $\vec{v}$. Since the force is perpendicular to the velocity,it does no work on the particle,and therefore,the speed of the particle remains constant inside the dees.
When the particle crosses the gap between the dees,it is subjected to an oscillating electric field. This electric field exerts an electric force on the particle,which does work on it,thereby increasing its kinetic energy and speed. Thus,the speed of the charged particle increases only in the gap between $D_1$ and $D_2$.

(C) When a charged particle is accelerated in a cyclotron,the radius $r$ of its path is given by $r = \frac{\sqrt{2Km}}{Bq}$.
Squaring both sides,we get $r^2 = \frac{2Km}{B^2q^2}$,which implies $K = \frac{B^2q^2r^2}{2m}$.
Since $B$ and $r$ are the same for all particles at the exit,$K \propto \frac{q^2}{m}$.
For protons $\left({ }_{1}^{1} H\right)$,deuterons $\left({ }_{1}^{2} H\right)$,and $\alpha$-particles $\left({ }_{2}^{4} He\right)$:
Charge ratio: $q_p : q_d : q_{\alpha} = 1 : 1 : 2$.
Mass ratio: $m_p : m_d : m_{\alpha} = 1 : 2 : 4$.
Calculating the ratio of kinetic energies $K \propto \frac{q^2}{m}$:
$K_p \propto \frac{1^2}{1} = 1$.
$K_d \propto \frac{1^2}{2} = 0.5$.
$K_{\alpha} \propto \frac{2^2}{4} = 1$.
Comparing the values,the minimum kinetic energy is gained by deuterons.

(B) In a cyclotron,the charged particle is accelerated by an electric field in the gap between the two dees.
Within the dees,the particle moves in a circular path due to the perpendicular magnetic field,where its speed remains constant.
Since the particle is constantly changing its direction of motion while moving in a circle,it experiences centripetal acceleration at all times.
Therefore,the particle undergoes acceleration throughout its motion.

(C) The cyclotron frequency is given by $f = \frac{qB}{2\pi m}$.
The maximum velocity $v$ of the proton at the radius $r$ is $v = \frac{qBr}{m} = 2\pi fr$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m(2\pi fr)^2 = 2\pi^2 mf^2r^2$.
Given values: $f = 10 \times 10^6 \text{ Hz}$, $r = 0.6 \text{ m}$, $m = 1.67 \times 10^{-27} \text{ kg}$.
Substituting these values:
$K = 2 \times (3.14)^2 \times (1.67 \times 10^{-27}) \times (10^7)^2 \times (0.6)^2$
$K = 2 \times 9.8596 \times 1.67 \times 10^{-27} \times 10^{14} \times 0.36$
$K \approx 1.185 \times 10^{-12} \text{ J}$.
To convert to $\text{MeV}$, divide by $1.6 \times 10^{-13} \text{ J/MeV}$:
$K = \frac{1.185 \times 10^{-12}}{1.6 \times 10^{-13}} \approx 7.4 \text{ MeV}$.
Rounding to the nearest given option, the kinetic energy is $7 \text{ MeV}$.

(C) The radius $(r)$ of the circular path of a charged particle moving perpendicular to a magnetic field $(B)$ is given by $r = \frac{mv}{qB}$.
From this relation,we can see that $r \propto v$,which means as the radius $(r)$ increases,the linear velocity $(v)$ of the particle must also increase.
The angular velocity $(\omega)$ is given by $\omega = \frac{v}{r}$.
Substituting $v = \frac{rqB}{m}$ into the expression for $\omega$,we get $\omega = \frac{rqB}{mr} = \frac{qB}{m}$.
Since the charge $(q)$,magnetic field $(B)$,and mass $(m)$ are constants,the angular velocity $(\omega)$ remains independent of the radius and velocity.
Therefore,as the radius increases,$v$ increases and $\omega$ remains constant.

(D) In a cyclotron,the radius $R$ of the path of an ion is given by the relation $R = \frac{mv}{qB}$,where $v$ is the velocity of the ion.
Rearranging this for velocity,we get $v = \frac{qBR}{m}$.
The kinetic energy $(K.E.)$ of the ion is given by the formula $K.E. = \frac{1}{2}mv^2$.
Substituting the expression for $v$ into the kinetic energy formula:
$K.E. = \frac{1}{2}m \left( \frac{qBR}{m} \right)^2$
$K.E. = \frac{1}{2}m \left( \frac{q^2B^2R^2}{m^2} \right)$
$K.E. = \frac{q^2B^2R^2}{2m}$.

(D) The cyclotron frequency is given by the formula $f = \frac{qB}{2\pi m}$.
Rearranging the formula to solve for the magnetic field $B$,we get $B = \frac{2\pi mf}{q}$.
Given values are:
Frequency $f = 20 MHz = 20 \times 10^6 Hz$
Charge of proton $q = 1.6 \times 10^{-19} C$
Mass of proton $m = 1.67 \times 10^{-27} kg$
Substituting these values into the formula:
$B = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 20 \times 10^6}{1.6 \times 10^{-19}}$
$B = \frac{209.536 \times 10^{-21}}{1.6 \times 10^{-19}}$
$B \approx 1.31 T$.

(A) The magnetic force acting on a charged particle moving in a magnetic field provides the necessary centripetal force for circular motion:
$F_m = F_c \Rightarrow qvB = \frac{mv^2}{r}$
From this, the radius of the path is $r = \frac{mv}{qB}$.
The time period of one revolution is $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$.
The frequency of rotation $f$ is the reciprocal of the time period:
$f = \frac{1}{T} = \frac{qB}{2\pi m}$.
Since $2\pi$ is a constant, the frequency is proportional to $\frac{qB}{m}$.