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I'm not fully grasping how $E=hv$ (Energy = Planck's constant $\times$ velocity) shows that the higher the wavelength, the lower the energy from the equation $c=v\lambda$ (speed of light = velocity $\times$ wavelength)

But $v=f\lambda$ (velocity = frequency $\times$ wavelength) so wouldn't that show that the higher the wavelength, the higher the energy? $E=hf\lambda$?

Winther
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hannah
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    1.) this question is terribly formated 2.) this is a physics question -> close – tired Jun 26 '16 at 22:36
  • The biggest mistake is pointed out in the answer below, but there are many other misconceptions here: 1) "speed of light = velocity × wavelength". Velocity of what exactly? If it's of photons then it would be $c$ and this equation would say that $\lambda = 1$. This is not true, what is true is that speed of light = wavelength × frequency. 2) Even if the equation was $E = hv$ with $v$ being the velocity of photons then the equation would say that all photons would have the same energy as they all have the same velocity. – Winther Jun 26 '16 at 23:51

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You misread a greek nu, $\nu$, for a Latin 'v'. The first equation is $E=h\nu$ with a nu, which denotes $f$.

(As Winther rightly pointed out in a comment, the second occurrence, in $c=\nu\lambda$, is also a nu that denotes the frequency $f$. Only the third occurrence, in $v=f\lambda$, is actually a $v$ for velocity.)

joriki
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