There is $$ E=\frac{h c}{\lambda} \tag{1} $$ and $$ E=\frac{h^2}{2 m \lambda^2} \,, \tag{2} $$ where $m$ is the mass of the object and $\lambda$ is the wavelength.
Can someone please tell me which one is to be used under what conditions? I am a bit confused. Thanks in advance!

  • 9
    $\begingroup$ Where did you find these formulas, respectively? What are $\lambda$ and $m$? What specifically is unclear to you about the text that must have accompanied them? $\endgroup$
    – ACuriousMind
    Jan 7, 2021 at 19:43
  • 1
    $\begingroup$ @ACuriousMind My guess is that they were found in an exam or in an assignment... $\endgroup$
    – user137289
    Jan 7, 2021 at 20:21

3 Answers 3


The De-Broglie approach tells us that the momentum of a wave is $$p=\frac{h}{\lambda}.$$ Thus for an electromagnetic wave ($m=0$, phase velocity $c$) the Energy is: $$E=pc=h\frac{c}{\lambda}=h\nu.$$ For a particle with mass $m$, which can also be described as a wave with wavelength $\lambda$ (e.g. electron) the kinetic Energy is calculated with: $$E_{\rm kin}=\frac{1}{2}mv^2=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}.$$

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    $\begingroup$ Probably worth adding that both equations can be traced to formulae for energy in terms of momentum. The energy of a photon is $E = pc$ and the kinetic energy of a massive particle is $E = \frac{p^2}{2m}$. Substituting $p = \frac{h}{\lambda}$ into both equations gives the desired "quantum" energy equations. $\endgroup$ Jan 8, 2021 at 5:54
  • $\begingroup$ So here, E=hc/λ is the totel energy of the particle? $\endgroup$ Jan 8, 2021 at 7:57
  • $\begingroup$ @ParamBudhadev yes because the Photon has no mass and no charge, so it can't have a potential Energy (in a gravitational or electric field). $\endgroup$
    – Roger
    Jan 8, 2021 at 10:28
  • The first one $E=\frac{hc}{\lambda}$ should be used for massless ($m=0$) particle.
  • The second one $E=\frac{h²}{2m\lambda²}$ should be used for massive ($m \neq 0$) particles.

The first equation gives the energy of a photon (zero mass) in terms of its wavelength, $\lambda$.

The second gives the kinetic energy of a particle of mass m (moving at a speed much less than $c$, the speed of light) in terms of its de Broglie wavelength, $\lambda$.


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