A proton is accelerating in a cyclotron where | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$(A)$ 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 revoluti

Vedclass · Accepted Answer

$100$

Vedclass · Answer

$(A)$ 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 $= 2 	imes e 	imes 10^5 \,V = 2 	imes 10^5 \,eV$.Target kinetic energy $E = 20 \,MeV = 20 	imes 10^6 \,eV = 2 	imes 10^7 \,eV$.Number of revolutions $n = \frac{E}{\Delta E} = \frac{2 	imes 10^7 \,eV}{2 	imes 10^5 \,eV} = 100$.

$A$ proton is accelerating in a cyclotron where the applied magnetic field is $2 \,T$. If the potential gap is effectively $100 \,kV$, then how many revolutions does the proton have to make between the "dees" to acquire a kinetic energy of $20 \,MeV$?

Similar Questions

$A$ cyclotron's oscillator frequency is $20 MHz$. The operating magnetic field for accelerating protons is (charge of proton $= 1.6 \times 10^{-19} C$,mass of proton $= 1.67 \times 10^{-27} kg$). (in $T$)

In the cyclotron, as the radius of the circular path of the charged particle increases, ($\omega$ = angular velocity, $V$ = linear velocity)

The oscillator frequency of a cyclotron is $10 \,MHz$. What should be the operating magnetic field for accelerating protons (in $,T$)? (Given: mass of proton $m = 1.67 \times 10^{-27} \,kg$,charge of proton $q = 1.6 \times 10^{-19} \,C$)

In a cyclotron,a charged particle

The maximum velocity to which a proton can be accelerated in a cyclotron of $10\, MHz$ frequency and radius $50\, cm$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module