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Question: A proton is accelerating on a cyclotron having oscillating frequency of 11MHz in external magnetic f...

A proton is accelerating on a cyclotron having oscillating frequency of 11MHz in external magnetic field of 1 T. If the radius of its dees is 55 cm, then its kinetic energy (in Me V) is

A

13.36

B

12.52

C

14.89

D

14.19

Answer

14.19

Explanation

Solution

Here, vc=11MHz=11×106 Hzv _ { c } = 11 \mathrm { MHz } = 11 \times 10 ^ { 6 } \mathrm {~Hz}

∴\therefore K.E. =q2 B2R22 m=(1.6×10−19)×(1)2×(55×10−2)22×1.67×10−27= \frac { \mathrm { q } ^ { 2 } \mathrm {~B} ^ { 2 } \mathrm { R } ^ { 2 } } { 2 \mathrm {~m} } = \frac { \left( 1.6 \times 10 ^ { - 19 } \right) \times ( 1 ) ^ { 2 } \times \left( 55 \times 10 ^ { - 2 } \right) ^ { 2 } } { 2 \times 1.67 \times 10 ^ { - 27 } } =23.19×10−13 J= 23.19 \times 10 ^ { - 13 } \mathrm {~J}

=23.19×10−131.6×10−19=14.49×106eV=14.49MeV= \frac { 23.19 \times 10 ^ { - 13 } } { 1.6 \times 10 ^ { - 19 } } = 14.49 \times 10 ^ { 6 } \mathrm { eV } = 14.49 \mathrm { MeV }