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(B) For a photon, the momentum $p$ is given by $p = \frac{h}{\lambda}$, where $h$ is Planck's constant.According to the de Broglie relation, the wavel

Vedclass · Accepted Answer

$\lambda' = \lambda$

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(B) For a photon, the momentum $p$ is given by $p = \frac{h}{\lambda}$, where $h$ is Planck's constant.According to the de Broglie relation, the wavelength $\lambda'$ associated with a particle of momentum $p$ is $\lambda' = \frac{h}{p}$.Substituting the expression for $p$ into the de Broglie equation, we get $\lambda' = \frac{h}{h/\lambda} = \lambda$.Therefore, the de Broglie wavelength of a photon is equal to the wavelength of the electromagnetic radiation associated with it.Thus, the correct option is $B$.

The relation between the wavelength of electromagnetic radiation $(\lambda)$ and the de Broglie wavelength of its quantum (photon) $(\lambda')$ is . . . . . . .

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