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Firstly with the help of $E= hf$ I calculated the energy by taking the frequency of light as $6 \times 10^{14}\ \rm Hz$(the range of frequency of light is from $4$ to $8\times10^{14}\ \rm Hz$ so I took $6$ as the mean). So I got the energy as $6.62607004\times 10^{-34} \times 6\times 10^{14}\ \rm kg\ m^2/s^2$.

Secondly we know that a photon has zero rest mass so the energy of a photon is calculated simply by $E = pc$ where momentum of a photon(red photon) is $9.816 \times 10^{-28}\ \rm kgm/s$, so by $c= \frac{E}{p}$ I got $c= 4.04\times 10^8\ \rm m/s$ which is clearly faster than speed of light. Tell me where am I wrong??

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    $\begingroup$ If you can't tell green from red, you probably shouldn't be driving. $\endgroup$
    – WillO
    Commented Jul 31, 2018 at 6:43

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The energy of a photon is indeed $h\nu$ and also $pc$ so we get:

$$ c = \frac{h\nu}{p} $$

The momentum of a photon is $h/\lambda$ and substituting this in the above gives:

$$ c = \nu\lambda $$

which is the well know expression for the velocity of a wave.

The problem with your calculation is that you're taking the frequency for green light and the wavelength for red light and combining them, and naturally that gives the wrong answer. If you multiply the wavelength of red light with the frequency of red light you will get the result $c$.

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