If the emitted radiation falls in the microwave region,the device is termed as

  • A
    $LASER$
  • B
    $MASER$
  • C
    Both $(a)$ and $(b)$
  • D
    None of these

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$A$ proton moving with a momentum $p_{1}$ has a kinetic energy $1/8$th of its rest mass-energy. Another light photon having energy equal to the kinetic energy of the proton possesses a momentum $p_{2}$. Then,the ratio $\frac{p_{1}-p_{2}}{p_{1}}$ is equal to

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If the energy of a photon is expressed in units of $KeV$ and the wavelength in units of $\mathring{A}$,then the energy of the photon can be calculated by:

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Reason: The photon behaves like a particle.

What is the charge on a photon?

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