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**__Solution:__
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A deuteron nucleus has the same charge as that of a proton and nearly twice
its mass.

Therefore, mass of deuteron,

and its charge,
.

In a cyclotron that has uniform magnetic field
,
the angular velocity of circular motion of a particle of charge
,
mass
is

and the corresponding frequency is

The final speed of the charged particles is determined by the radius *R*
at which the particles leave the accelerator. The corresponding (nonrelativistic
kinetic energy) of the particles is

The oscillator frequency of the cyclotron is

.

Therefore, the final kinetic energy of the deuterons will be

The accelerator potential across the dees is 80 kV. Therefore, kinetic energy
of the deuterons will increase each time they cross the dee-gap by an amount

Therefore, the total number of accelerations that a deuteron will receive
during its acceleration in the cyclotron will be

The kinetic energy of the deuteron, after undergoing acceleration each time
will change by
. That is
it will change from

to

And, the speed of the deuteron will change from

to
to

With constant speeds deuterons will travel for time

Therefore, the total distance a deuteron will travel before leaving the
cyclotron will be

*The reader is encouraged to evaluate the sum of square roots of natural numbers ranging from 1 to 208 and see the resultant value...*
,

Therefore,