4.5 x \(10^{10}\)Hz
5 x \(10^{9}Hz\)
4.5 x \(10^{8}Hz\)
2 x \(10^{7}Hz\)
2 x \(10^{6}Hz\)
Correct answer is E
V=f\(\lambda, f =\frac{v}{\lambda}= \frac{3\times10^8}{150}\Rightarrow 2\times10^6\)Hz
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