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 much revolutions the proton has to make between the "dees" to acquire a kinetic energy of 20 MeV ?

Question

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 much revolutions the proton has to make between the "dees" to acquire a kinetic energy of 20 MeV ? (A) 100 (B) 150 (C) 200 (D) 300

Tardigrade Admin · Accepted Answer

Energy increased in each revolution =2 × 100 × 103 eV =2 × 105 eV Now for energy E=2 × 107 eV Number of revolution =(2 × 107 eV /2 × 105 eV )=100

Q. 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 much revolutions the proton has to make between the "dees" to acquire a kinetic energy of $20 M e V ?$

100

150

200

300

Energy increased in each revolution $= 2 \times 100 \times 1 0^{3} e V$
$= 2 \times 1 0^{5} e V$
Now for energy $E = 2 \times 1 0^{7} e V$
Number of revolution $= \frac{2 \times 1 0 ^{7} e V}{2 \times 1 0 ^{5} e V} = 100$

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

100

150

200

300

Energy increased in each revolution =2×100×103eV =2×105eV Now for energy E=2×107eV Number of revolution =2×105eV2×107eV​=100

Q. 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 much revolutions the proton has to make between the "dees" to acquire a kinetic energy of $20 M e V ?$

Energy increased in each revolution $= 2 \times 100 \times 1 0^{3} e V$
$= 2 \times 1 0^{5} e V$
Now for energy $E = 2 \times 1 0^{7} e V$
Number of revolution $= \frac{2 \times 1 0 ^{7} e V}{2 \times 1 0 ^{5} e V} = 100$