# In the photoelectric effect, why we use momentum to calculate kinetic energy?

If we know the wavelength of an ejected electron, why do we use the momentum equation ($E_\mathrm k=\frac{p^2}{2m}$, $\lambda = h/p$). Why can't we use $\frac{1}{2}mv^2$ instead, using $c/\lambda = v$?

Does it have something to do with particle-like or wave-like behavior?

• Try using mv^2/2 and see what happens. Commented Jan 15, 2017 at 7:32
• Also note that $c/\lambda=f$ where $f$ is frequency (not velocity).
– user7951
Commented Jan 15, 2017 at 10:37

# Relativity

Since electrons are travelling at near the speed of light, we need to use relativity.

In relativity, the correct equation relating energy to mass and momentum is the energy-momentum relation:

$$E^2 = (m_0 c^2)^2 + (p c)^2$$

where:

• $$E$$ denotes the total energy of the particle.
• $$m_0$$ denotes the (rest) mass of the particle.
• $$c$$ denotes the speed of light.
• $$p$$ denotes the momentum of the particle.

The momentum of a particle can be found from its (rest) mass and its velocity:

$$\displaystyle p = \gamma m_0 v$$

where:

$$\displaystyle \gamma = \frac {1} {\sqrt{1 - \left({ \frac v c }\right)^2}} = \left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5}$$

Note that $$p = m v$$ is only true when $$0 \lt v \ll c$$, when $$\gamma \approx 1$$.

# Kinetic energy

When the particle is not moving, $$p = 0$$, so $$E = m_0 c^2$$ (mass-energy equivalence) is denoted the rest energy.

Substituting the definition of momentum gives:

$$\displaystyle E^2 = m_0^2 c^4 + \gamma^2 m_0^2 v^2 c^2 = m_0^2 c^4 \left({\frac {c^2} {c^2 - v^2}}\right) = m_0^2 c^4 \gamma^2$$

Now, the kinetic energy is defined as the total energy minus the rest energy:

$$E_{\mathrm k} = E - m_0 c^2$$

$$\displaystyle E_{\mathrm k} = m_0 c^2 \left({\gamma - 1}\right)$$

# Classical approximation: $$\displaystyle E_\mathrm k = \frac 12 m_0 v^2$$

Now, the classical formula $$\displaystyle E_\mathrm k = \frac 12 m_0 v^2$$ is an approximation of $$E_\mathrm k$$ when $$0 \lt v \ll c$$:

$$E_{\mathrm k}$$

$$\displaystyle = m_0 c^2 \left({\gamma - 1}\right)$$

$$\displaystyle = m_0 c^2 \left({\left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5} - 1}\right)$$

$$\displaystyle \approx m_0 c^2 \left({1 - (-0.5) \left({ \frac v c }\right)^2 - 1}\right)$$

$$\displaystyle = \frac 1 2 m_0 v^2$$

However, as I have mentioned, it only works when $$v$$ is very small, which is not the case for electrons.

# Another approximation: $$\displaystyle E_{\mathrm k} = \frac {p^2} {2 m_0}$$

$$E_{\mathrm k}$$

$$\displaystyle = \sqrt{m_0^2 c^4 + p^2 c^2} - m_0 c^2$$

$$\displaystyle = m_0 c^2 \sqrt{1 + \frac {p^2} {m_0^2 c^2}} - m_0 c^2$$

$$\displaystyle \approx m_0 c^2 \left({1 + \frac {p^2} {2 m_0^2 c^2}}\right) - m_0 c^2$$

$$\displaystyle = \frac {p^2} {2 m}$$

# Comparison

Now, for reasons unknown, the second approximation is actually better than the first:

(For convenience reasons, $$c$$ is took to be $$1$$ in the graph. It can be seen that the second approximation (blue) goes to infinity with the real kinetic energy (black) as $$v \to c$$, while the first approximation (red) does not.)

Perhaps this is the reason why the second approximation is used instead of the first.

Note that it is still an approximation.

Another reason is that, $$\displaystyle \nu = \frac c \lambda$$, but that is $$\nu$$ (Greek letter nu) for frequency not the $$v$$ (Latin letter vee) for velocity. The former has a unit of $$\mathrm{s^{-1}}$$ while the latter has a unit of $$\mathrm{m~s^{-1}}$$.

• oh so..actually it is nothing to do with velocity...and we can't even get the velocity directly from wavelength. And if I want to use energy-momentum equation, that would be very complicated. Therefore, we use Ek=p^2/2m. Did i understand correctly? Commented Jan 16, 2017 at 5:35
• Actually it doesn't seem that complicated to me though. Commented Jan 16, 2017 at 5:43
• @pronoob reasons for using p^2/2m added.
– DHMO
Commented Jan 16, 2017 at 11:44

It's just because photon in newtonian frame of reference us assumed to be having no mass at rest.....so the mass factor in momentum is undefined in case of photon so in order to measure momentum we use kinetic energy. For electron too due to reason that its mass increase on speed so above method more convenient.