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Step-by-Step Solution
Step 1: State the energy-gain principle in the cyclotron
In a cyclotron, the charged particle (proton) gains energy each time it crosses the gap between the dees. Since the proton crosses the gap twice per revolution, the total energy gained per revolution is
2\,e\,V,
where
e
is the charge on the proton and
V
is the accelerating potential provided by the radio frequency oscillator.
Step 2: Write down the expression for kinetic energy
When the proton reaches a speed
v,
its kinetic energy is
\frac{1}{2} m_p v^2,
where
m_p
is the mass of the proton. The total energy gained by the proton after
n
revolutions must be equal to its final kinetic energy:
n \bigl(2\,e\,V\bigr) = \frac{1}{2} \, m_p \, v^2.
Step 3: Substitute the known values
From the problem,
V = 12\,\text{kV} = 12 \times 10^3\,\text{V},
m_p = 1.67 \times 10^{-27}\,\text{kg},
e = 1.6 \times 10^{-19}\,\text{C},
and the proton eventually attains
\frac{1}{6}
of the speed of light
c = 3 \times 10^8\,\text{m/s}.
So,
v = \frac{c}{6} = \frac{3 \times 10^8}{6} = 5 \times 10^7\,\text{m/s}.
Hence, the equation becomes:
n \Bigl(2 \times 1.6 \times 10^{-19}\,\text{C} \times 12 \times 10^3\,\text{V}\Bigr) \;=\; \frac{1}{2} \times 1.67 \times 10^{-27}\,\text{kg} \times \bigl(5 \times 10^7\,\text{m/s}\bigr)^2.
Step 4: Perform the numerical calculation
First, calculate the left-hand side coefficient (energy gain per revolution):
2 \times 1.6 \times 10^{-19} \times 12 \times 10^3
= 2 \times 1.6 \times 12 \times 10^{-19} \times 10^3
= 38.4 \times 10^{-16}\,\text{J}
= 3.84 \times 10^{-15}\,\text{J}.
Then, compute the right-hand side:
\frac{1}{2} \times 1.67 \times 10^{-27} \times (5 \times 10^7)^2
= 0.5 \times 1.67 \times 10^{-27} \times 25 \times 10^{14}
= 0.5 \times 1.67 \times 25 \times 10^{-27+14}
= 0.5 \times 1.67 \times 25 \times 10^{-13}\,\text{J}.
Let's break that down:
1.67 \times 25 = 41.75.
0.5 \times 41.75 = 20.875.
20.875 \times 10^{-13}\,\text{J} = 2.0875 \times 10^{-12}\,\text{J}.
Step 5: Solve for the number of revolutions n
So the equation
n \times 3.84 \times 10^{-15}\,\text{J} = 2.0875 \times 10^{-12}\,\text{J}
gives
n = \frac{2.0875 \times 10^{-12}}{3.84 \times 10^{-15}}
= \frac{2.0875}{3.84} \times 10^{-12 + 15}
= 0.543 \times 10^3
= 543.4 \,(\text{approximately}).
Thus, the proton makes about
\boxed{543}
revolutions.
