Two opposite charged particles oscillate about their mean equilibrium position in free space, with a frequency of $10^{9} H z$ . The wavelength of the corresponding electromagnetic wave produced is ______

0.3 m

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10^{9} m

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3 \times 10^{17} m

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3.3 m

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Solution

The correct option is B $0.3 m$
Electromagnetic wave $V = 3 \times 10^{8} m / s$
Given frequency $(f) = 10^{9} H z$
$V = f λ λ = \frac{V}{f} = \frac{3 \times 10^{8}}{10^{9}} = 0.3 m$

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Two opposite charged particles oscillate about their mean equilibrium position in free space, with a frequency of 109Hz. The wavelength of the corresponding electromagnetic wave produced is ______

The correct option is B 0.3mElectromagnetic wave V=3×108m/sGiven frequency (f)=109HzV=fλλ=Vf=3×108109=0.3m

Two opposite charged particles oscillate about their mean equilibrium position in free space, with a frequency of $10^{9} H z$ . The wavelength of the corresponding electromagnetic wave produced is ______

The correct option is B $0.3 m$
Electromagnetic wave $V = 3 \times 10^{8} m / s$
Given frequency $(f) = 10^{9} H z$
$V = f λ λ = \frac{V}{f} = \frac{3 \times 10^{8}}{10^{9}} = 0.3 m$