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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 waveleng

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$\lambda' = \lambda$

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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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