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(C) The energy of a photon is given by $E = h
u = \frac{hc}{\lambda}$.The momentum of a photon is given by $p = \frac{h}{\lambda}$.By dividing these

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Increases by a factor of $4$

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(C) The energy of a photon is given by $E = h
u = \frac{hc}{\lambda}$.The momentum of a photon is given by $p = \frac{h}{\lambda}$.By dividing these two equations,we get $E = pc$,where $c$ is the speed of light.Since $c$ is a constant,the momentum $p$ is directly proportional to the energy $E$ $(p \propto E)$.If the energy $E$ is increased by a factor of $4$,the momentum $p$ will also increase by a factor of $4$.

If the energy of the photon is increased by a factor of $4$,then its momentum

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