# How does the equation $h\nu = KE + W$ account for the inability of electrons to exceed the speed of light?

To start, obviously, electrons cannot surpass the speed of light; however, the equation describing the kinetics of an electron hit with light with a higher energy photon than the work function, $$W$$, makes it seem as if an electron could exceed the speed of light.

Taking a caesium atom for example, which has a work function of $$3.42\times10^{-19}$$ J, and solving for the wavelength of light required to make an electron reach the speed of light shows me that gamma radiation should allow an electron to meet the speed of light.

$$h\nu = KE + W\tag{1}$$

$$KE=\frac{1}{2}m_ev^2\tag{2}$$

$$6.626\times10^{-34}\times\nu=\frac{1}{2}\times9.1094\times10^{-31}(2.998\times10^{8})^2\tag{3}$$

$$\nu = 6.18 \times 10^{19}\tag{4}$$

$$\lambda=\frac{c}{\nu}\tag{5}$$

$$\lambda=\frac{2.998\times10^{8}}{6.18\times10^{-19}}\tag{6}$$

$$\lambda=4.85\times10^{-3}\text{ (this is the wavelength of gamma radiation)}\tag{7}$$

Does this equation not work for all frequencies of electromagnetic radiation, if so what is the explanation?

The equation KE$$=mv^2/2$$ only holds for non-relativistic particles moving at velocities far lower than the speed of light. It implies that increasing the kinetic energy without bound would also increase the velocity without bound, which we know gets corrected by relativity. You need $$E=\sqrt{p^2c^2+m^2c^4}, \quad v=\frac{pc^2}{E}$$ to describe relativistic electrons. With this, you can show that there is no photon frequency that would cause the electron to accelerate beyond the speed of light.