Chapter 26, Problem 28P

(a)

To determine

The work done to accelerate a proton.

Answer to Problem 28P

  $\text{2}\text{.23}\times {\text{10}}^{-9}\text{J}$ .

Explanation of Solution

Given:

Speed, $v=0.997c$

Formula used:

It is known that:

  $W=\Delta KE=\left(m-m_0\right)c^2=m_0{c}^2\left[\left\{\frac{1}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}\right\}-1\right]$

Where,

  $m_0$ is rest mass

m is relativistic mass

v is velocity

c is a speed of light, $3.00\times {10}^{8}m/s$

Calculation:

It is known that the mass of a proton is $=1.67\times {10}^{-27}\text{kg}$

By substituting the values in the formula:

  $W\text{=1}\text{.67}\times {\text{10}}^{-27}\times {\left(3\times {10}^8\right)}^2\left[\left\{\frac{\text{1}}{{\left[\text{1}-{\left(\text{0}\text{.997}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}}\right\}-\text{1}\right]$

  $\text{=1}\text{.79}\times {\text{10}}^{-9}\text{J}$

(b)

To determine

The momentum of proton.

Answer to Problem 28P

  $\text{6}\text{.46}{\text{×10}}^{\text{-18}}\text{kg}\cdot \text{m/s}$ .

Explanation of Solution

Given:

Speed, $v=0.997c$

Formula used:

It is known that the momentum of proton:

  $p=\frac{m_{0}v}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

Where,

  $m_0$ is rest mass

v is velocity

c is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

Substitute the given values in the formula $p=\frac{m_{0}v}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

  $p=\frac{\left(1.67\times {10}^{-27}\text{kg}\right)\left(0.997\right)\left(3.00\times {10}^8\text{m/s}\right)}{{\left[1-{\left(0.997\right)}^2\right]}^{\frac{1}{2}}}$ $=6.46\times {10}^{-18}kg\cdot m/s$