The magnitude of velocity of an electron in the laboratory frame.

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Answer 1

Question

Chapter 39, Problem 41PQ

To determine

The magnitude of velocity of an electron in the laboratory frame.

Expert Solution & Answer

Answer to Problem 41PQ

The magnitude of velocity of the electron measured in the laboratory frame is $8.50×107 m/s$ .

Explanation of Solution

Write the expression for the magnitude of velocity measured from the laboratory frame.

$v=vx2+vy2+vz2$ (I)

Here, $v$ is the resultant velocity, $vx$ is the velocity transformation in the $x$ direction, $vy$ is the velocity transformation in the $y$ direction, and $vz$ the velocity transformation in the $z$ direction.

Write the expression for velocity components in each direction,

$vx=vx′+vrel[1+vrel×vx′c2]$ (II)

$vy=vy′γ[1+vrel×vx′c2]$ (III)

$vz=vz′γ[1+vrel×vx′c2]$ (IV)

Here, $vrel$ is the relative velocity, $c$ is the speed of light, $γ$ is the Lorentz constant, $vx′$ is the inverse velocity transformation in $x$ direction, $vy′$ is the inverse velocity transformation in $y$ direction and $vz′$ is the inverse velocity transformation in $z$ direction.

Write the expression for Lorentz constant,

$γ=11−(vrelc)2$ (V)

Conclusion:

Substitute $2.00×107 m/s$ for $vrel$ and $3.00×108m/s$ for $c$ in equation (V) to find $γ$ .

$γ=11−(2.00×107m/s3.00×108m/s)2=11−(0.06)2=10.998=1.002$

Substitute $5.00×107 m/s$ for $vx′$ , $2.00×107 m/s$ for $vrel$ and $3.00×108m/s$ for $c$ in equation (III) to find $vx$ .

$vx=5.00×107 m/s+2.00×107 m/s[1+5.00×107 m/s×2.00×107 m/s(3.00×108)2m2/s2]=7.00×107 m/s[1+10159×1016]=7.00×107m/s1.011=6.92×107 m/s$

Substitute $4.00×107 m/s$ for $vy′$ , $1.002$ for $γ$ , $5.00×107 m/s$ for $vx′$ , $2.00×107 m/s$ for $vrel$ and $3.00×108m/s$ for $c$ in equation (IV) to find $vy$ .

$vy=4.00×107 m/s1.002[1+2.00×107 m/s×5.00×107 m/s(3.00×108)2 m2/s2]=4.00×107 m/s1.002[1.011]=4.00×107m/s1.013=3.95×107 m/s$

Substitute $3.00×107 m/s$ for $vz′$ , $1.002$ for $γ$ , $5.00×107 m/s$ for $vx′$ , $2.00×107 m/s$ for $vrel$ and $3.00×108m/s$ for $c$ in equation (V) to find $vz$ .

$vz=3.00×107 m/s1.002[1+2.00×107 m/s×5.00×107 m/s(3.00×108)2 m2/s2]=3.00×107 m/s1.002[1.011]=3.00×1071.013=2.96×107 m/s$

Substitute $6.92×107 m/s$ for $vx$ , $3.95×107 m/s$ for $vy$ and $2.96×107 m/s$ for $vz$ in the equation (I) to obtain the velocity of the electron.

$v=vx2+vy2+vz2=(6.92×107 m/s)2+(3.95×107 m/s)2+(2.96×107 m/s)2=72.2505×1014 m2/s2=8.50×107m/s$

Therefore, the magnitude of velocity of electron is $8.50×107m/s$ .

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