Question:
If the wavelength of an electromagnetic radiation is 4288 \(\AA\), then the de Broglie wavelength associated with its photon is:
If the wavelength of an electromagnetic radiation is 4288 \(\AA\), then the de Broglie wavelength associated with its photon is:
Show Hint
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
Solution and Explanation
Concept:
For a photon,
\[
p=\frac{h}{\lambda}.
\]
De-Broglie wavelength is
\[
\lambda_d=\frac{h}{p}.
\]
Step 1: Substitute momentum of photon.
\[ \lambda_d = \frac{h}{h/\lambda} \] \[ =\lambda. \]
Step 2: Use given wavelength.
\[ \lambda=4288\AA. \] Hence, \[ \boxed{4288\AA} \]
Step 1: Substitute momentum of photon.
\[ \lambda_d = \frac{h}{h/\lambda} \] \[ =\lambda. \]
Step 2: Use given wavelength.
\[ \lambda=4288\AA. \] Hence, \[ \boxed{4288\AA} \]
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