Question:medium

If the wavelength of an electromagnetic radiation is 4288 \(\AA\), then the de Broglie wavelength associated with its photon is:

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Every photon automatically satisfies \[ \lambda_{de\,Broglie} = \lambda_{EM}. \]

Updated On: Jun 18, 2026

\(4288 \AA\)
\(1072 \AA\)
\(2144 \AA\)
\(8576 \AA\)

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The Correct Option is A

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