Chapter 6, Problem 60P

(a)

To determine

The average power delivered to the protons.

Explanation of Solution

Given:

Each of the proton has a kinetic energy $10\text{keV}$ . Per second $1\times {10}^9$ protons are reaching the target at the end of the $1.50\text{m}$ long accelerator. Mass of the proton is $1.67\times {10}^{-27}\text{Kg}$ .

Formula Used:

Write the expression for the power of the protons required to reach the end of the accelerator.

  $P\text{}=n\left(\text{Energy}\text{of}\text{each}\text{proton}\right)$

Here, $P$ is the required power and $n$ is the number of protons reaching the target per second.

Calculation:

Substitute $1\times {10}^9$ for $n$ and $10\text{keV}$ for energy of each protonin the above expression.

  $\begin{array}{c}P=\left(1\times {10}^9{\text{s}}^{\text{-1}}\right)\left(10\text{keV}\right)\\ \text{=}\left({\text{10}}^9{\text{s}}^{\text{-1}}\right)\left(\frac{\left(10\text{keV}\right)\left(1000\text{eV}\right)}{1\text{keV}}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)\\ =1.6\times {10}^{-6}\text{J}{\text{s}}^{\text{-1}}\\ =1.6\times {10}^{-6}\text{W}\end{array}$

Conclusion:

Thus, the average power that must be delivered to the stream of protons is $1.6\times {10}^{-6}\text{W}$ .

(b)

To determine

The force acting on the protons.

Explanation of Solution

Given:

Each of the proton has a kinetic energy $10\text{keV}$ . Per second $1\times {10}^9$ protons are reaching the target at the end of the $1.50\text{m}$ long accelerator. Mass of the proton is $1.67\times {10}^{-27}\text{Kg}$ .

Formula Used:

The protons will perform some work at the expense of the energy with which they are emitting from the accelerator.

Write the expression for the work done by the protons.

  $W=(\text{Energy}\text{of}\text{each}\text{proton})t$ …… (1)

Here, $W$ is the work done and $t$ is the time required to eject one proton.

Write the expression for the net force on the stream protons.

  $F_{net}=\frac{W}{S}$ …… (2)

Here, $W$ is the work done, $F_{\text{net}}$ is the net force and $S$ is the displacement.

Write the expression for the force on each proton.

  $F=\frac{F_{\text{net}}}{\left(\text{number}\text{of}\text{protons}\right)}$ …… (3)

Calculation:

Substitute $1\text{s}$ for $t$ and $1.6\times {10}^{-6}\text{W}$ for energy of eachproton in expression(1).

  $\begin{array}{c}W\text{=}\left(1.6\text{}\times {10}^{-6}\text{W}\right)\left(1\text{s}\right)\\ =1.6\text{}\times {10}^{-6}\text{J}\end{array}$

Because of the application of this constant force on the protons, they reach to the end of the accelerator by traveling a distance 1.50 m.

Substitute $1.6\times {10}^{-6}\text{W}$ for $W$ and $1.50\text{m}$ for S in expression (2).

  $\begin{array}{c}{F}_{\text{net}}=\frac{\text{1}\text{.6}\times {\text{10}}^{-6}\text{J}}{1.50\text{m}}\\ =1.07\times {10}^{-6}\text{N}\end{array}$

Substitute $1\times {10}^9$ for the number of protons in expression (3).

  $\begin{array}{c}F=\frac{1.07\times {10}^{-6}\text{N}}{1\times {10}^9}\\ =1.07\times {10}^{-15}\text{N}\end{array}$

Conclusion:

Thus, the force applied to each protons is $1.07\times {10}^{-15}\text{N}$ .

(c)

To determine

The velocity of protons just before hitting the target.

Explanation of Solution

Given:

Each of the proton has a kinetic energy $10\text{keV}$ . Per second $1\times {10}^9$ protons are reaching the target at the end of the $1.50\text{m}$ long accelerator. Mass of the proton is $1.67\times {10}^{-27}\text{Kg}$ .

Formula Used:

Write the expression for the acceleration of each proton.

  $a=\frac{F}{m_p}$ …… (4)

Here, $a$ is the acceleration and $m_p$ is the mass of each proton.

Write the expression for the final velocity of the protons just before hitting the target.

  $v_f^2=2aS$ …… (5)

Here, $v_f$ is the final velocity and $S$ is the displacement.

Calculation:

Substitute $1.07\times {10}^{-15}\text{N}$ for $F$ and $1.67\times {10}^{-27}\text{Kg}$ for $m_p$ in expression (4).

  $\begin{array}{c}a=\frac{1.07\times {10}^{-15}\text{N}}{1.67\times {10}^{-27}\text{Kg}}\\ =6.41\times {10}^{11}\text{m}{\text{s}}^{\text{-2}}\end{array}$

Substitute $6.41\times {10}^{11}\text{m}{\text{s}}^{\text{-2}}$ for $a$ and $1.5\text{m}$ for S in expression (5).

  $\begin{array}{c}{v}_f^2=2\left(6.41\times {10}^{11}\text{m}\text{}{\text{s}}^{\text{-2}}\right)\left(1.5\text{m}\right)\\ \text{=}\text{1}\text{.923}\text{}\times {10}^{12}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\end{array}$

Simplify the above expression.

  $\begin{array}{c}{\text{v}}_f=\sqrt{\text{1}\text{.923}\text{}\times {10}^{12}}\text{m}{\text{s}}^{\text{-1}}\\ =1.38\times {10}^6\text{m}{\text{s}}^{\text{-1}}\end{array}$

Conclusion:

Thus, the speed of the protons just before hitting the target is $1.38\times {10}^6\text{m}{\text{s}}^{\text{-1}}$ .