
A proton is accelerating on a cyclotron having an oscillating frequency of 11MHz in an external magnetic field of 1T. If the radius of its dees is 55 cm, then its kinetic energy (in MeV) is ($m_p=1.67\times 10^{-27}kg,e=1.6\times 10^{-19}C$):
A. 13.36
B. 12.52
C. 14.89
D. 14.49
Answer
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Hint: First of all it’s important to understand what is cyclotron and why it is used. Cyclotron is a device or setup used for accelerating the charged particle in order to achieve a very high speed. For physics experiments, like quantum physics, we need to understand the properties of charges or matter at very high velocities. This is the main reason why cyclotron is used.
Formula used:
$K.E. = \dfrac 12 mv^2, r=\dfrac{mv}{Bq}$
Complete answer:
Given the radius of Dees, i.e. the radius of area in which the proton will rotate = 55cm or 0.55 m.
B=1 Tesla, this field is applied in a direction perpendicular to Dees.
Frequency of cyclotron ($\nu$) = 11MHz
Now, for kinetic energy, we know
$K.E. = \dfrac 12 mv^2$
$And, r=\dfrac{mv}{Bq}$
$or\ v=\dfrac{Brq}m$
Hence, $K.E. = \dfrac 12 m\left ( \dfrac{Brq}m \right)^2$
Or $K.E. = \dfrac {B^2r^2q^2}{2m}$
Now, putting the values in the above expression, we get
$K.E. = \dfrac{1^2\times (0.55)^2 (1.6\times 10^{-19})^2}{2\times 1.67\times 10^{-27}} = 0.2319 \times 10^{-11} J$
Now, as $1 J=\dfrac1{1.6\times 10^{-19}} eV$
Hence $K.E. = \dfrac{0.2319\times 10^{-11}}{1.6\times 10^{-19}}eV = 14.49\times 10^6 eV= 14.49 MeV$
So, the correct answer is “Option D”.
Note:
It is important to note that the given frequency is of the changing polarity of the source attached to Dees for the generation of electric field. Electric field is important because the magnetic field can’t alter the speed of the particle. Its role is just to keep protons moving in circular motion. It is the electric field that increases the speed of the charge. The Dees of cyclotron uses alternating current for reversing the polarity so as to accelerate the charge efficiently. The cyclotron frequency is well matched with the frequency of charged particles.
Formula used:
$K.E. = \dfrac 12 mv^2, r=\dfrac{mv}{Bq}$
Complete answer:
Given the radius of Dees, i.e. the radius of area in which the proton will rotate = 55cm or 0.55 m.
B=1 Tesla, this field is applied in a direction perpendicular to Dees.
Frequency of cyclotron ($\nu$) = 11MHz
Now, for kinetic energy, we know
$K.E. = \dfrac 12 mv^2$
$And, r=\dfrac{mv}{Bq}$
$or\ v=\dfrac{Brq}m$
Hence, $K.E. = \dfrac 12 m\left ( \dfrac{Brq}m \right)^2$
Or $K.E. = \dfrac {B^2r^2q^2}{2m}$
Now, putting the values in the above expression, we get
$K.E. = \dfrac{1^2\times (0.55)^2 (1.6\times 10^{-19})^2}{2\times 1.67\times 10^{-27}} = 0.2319 \times 10^{-11} J$
Now, as $1 J=\dfrac1{1.6\times 10^{-19}} eV$
Hence $K.E. = \dfrac{0.2319\times 10^{-11}}{1.6\times 10^{-19}}eV = 14.49\times 10^6 eV= 14.49 MeV$
So, the correct answer is “Option D”.
Note:
It is important to note that the given frequency is of the changing polarity of the source attached to Dees for the generation of electric field. Electric field is important because the magnetic field can’t alter the speed of the particle. Its role is just to keep protons moving in circular motion. It is the electric field that increases the speed of the charge. The Dees of cyclotron uses alternating current for reversing the polarity so as to accelerate the charge efficiently. The cyclotron frequency is well matched with the frequency of charged particles.
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