Chapter 26, Problem 51GP

(a)

To determine

The speed of an electron.

Answer to Problem 51GP

  $\text{0}\text{.77}\text{m/s}$ .

Explanation of Solution

Given:

Kinetic energy is $\text{14,000}$ times its rest energy.

Formula used:

It is known that,

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

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

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

Calculation:

Calculate the speed from $m=\frac{m_0}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

  $14,000m_0{c}^2=\frac{m_0{c}^2}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-m_0{c}^2$

This gives ${\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}=7.14\times {10}^{-5}$

Or ${\left(\frac{v}{c}\right)}^2=1-5.10\times {10}^{-9}$

Now, take the square root,

  $\frac{v}{c}={\left(1-5.10\times {10}^{-9}\right)}^{\frac{1}{2}}\approx 1-\frac{1}{2}\left(5.10\times {10}^{-9}\right)=1-2.55\times {10}^{-9}$

The speed is $\text{0}\text{.77}\text{m/s}$ less than c

(b)

To determine

The contracted length of the tube.

Answer to Problem 51GP

  $\text{21}\text{cm}$ .

Explanation of Solution

Given:

Kinetic energy is $\text{14,000}$ times its rest energy.

The speed is $\text{0}\text{.77}\text{m/s}\text{less than c}$

Distance is $\text{3}\text{.0}\text{km}$

Formula used:

It is known that,

  $L=L_0{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}$

Where,

  $L_0$ is rest length

  $L$ is length (contracted)

  $v$ is velocity

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

Calculation:

Substitute the given values in the formula $L=L_0{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}$

  $=\left(3.0km\right)\left(7.14\times {10}^{-5}\right)=2.1\times {10}^{-4}km=21cm$