# Two different formulas for the calculation of energy in QM

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!

• 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? Jan 7, 2021 at 19:43
• @ACuriousMind My guess is that they were found in an exam or in an assignment...
– user137289
Jan 7, 2021 at 20:21

## 3 Answers

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: $$E_{\rm kin}=\frac{1}{2}mv^2=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}.$$

• 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. Jan 8, 2021 at 5:54
• So here, E=hc/λ is the totel energy of the particle? Jan 8, 2021 at 7:57
• @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). 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, $$\lambda$$.