Chapter 31: Q. 21 (page 901)

A $200 M W$ laser pulse is focused with a lens to a diameter of $2.0 m m$ .
a. What is the laser beam’s electric field amplitude at the focal point?
b. What is the ratio of the laser beam’s electric field to the electric field that keeps the electron bound to the proton of a hydrogen atom? The radius of the electron orbit is $0.053 n m$ .

Short Answer

Expert verified

Ratio of the laser beams electric field to the electric field that keeps electrons $= 2 \times 10^{- 2}$

Step by step solution

Laser beam’s electric field amplitude at the focal point

The amplitude of electromagnetic waves refers to the electric and magnetic fields' maximal field strength. The wave amplitude determines the wave energy. The amount of energy transported by a wave is determined by its amplitude.

Find Electric field

$P = 200 \times 10^{6} W$

$r = 2 \times 10^{- 6} m$

$A = π {(2 \times 10^{- 6})}^{2}$

$A_{v g}$ Intensity:

I_{o} = \frac{P}{A}

$I_{o} = \frac{200 \times 10^{6}}{π {(2 \times 10^{- 6})}^{2}}$

Electric field:

$E_{o} = {(\frac{2 I_{a v g}}{c \cdot ε_{o}})}^{1 / 2}$

$E_{o} = {(\frac{2 \times 15.9 \times 10^{18}}{3 \times 10^{8} \times 8.85 \times 10^{- 12}})}^{1 / 2}$

$E_{o} = 1 \times 10^{11} V / m$

Find the ratio

Electric field between electron and proton:

, $E = \frac{1}{4 π ε_{o}} \times \frac{e}{r^{2}}$

$E = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 18}}{{(0.053 \times 10^{- 9})}^{2}}$

$E = 5.12 \times 10^{12}$

Ratio:

$\frac{E_{o}}{E} = \frac{1 \times 10^{11}}{5.12 \times 10^{12}} = 2 \times 10^{- 2}$

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Laser beam’s electric field amplitude at the focal point

Find Electric field

Find the ratio

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