Light of wave length590nm is barely able to eject electrons from a metal plate. What would be the speed of the fastest electrons ejected by the light of one-third the wavelength?

$\varphi =\text{3.676}\times {\text{10}}^{\text{- 19}}\text{J}$

In the photoelectric effect, where light is evident in the form of particles and ejection of electrons takes place, the wavelength of incident light is either equal or below the threshold wavelength determined by the metal work function.

When the energy of incident light equals the work function, the energy is barely enough to eject electrons. Hence, the work function corresponds to an EM wave having a wavelength of $590\text{nm}$.

$$\begin{gathered} \varphi =h\frac{c}{\lambda } \\ \varphi =6.626\times {10}^{-34}\text{J}·\text{s}\times \frac{2.998\times {10}^8\text{m}·\text{s}}{590\text{nm}} \\ \varphi =6.626\times {10}^{-34}\text{J}.\text{s}\times \frac{2.998\times {10}^8\text{m}.{\text{s}}^{-1}}{590\times {10}^{-9}\text{m}} \\ \varphi =3.367\times {10}^{-19}\text{J} \end{gathered}$$

When the wavelength $=\frac{590\text{nm}}{3}$,

Applying,

$KE=hf-\varphi$

$KE=h\frac{c}{\lambda }-\varphi$

$\begin{array}{rcl}KE& =& 6.626\times {10}^{-34}\text{J}·\text{s}\times \frac{2.998\times {10}^8\text{m}·\text{s}}{\left(\frac{590\text{nm}}{3}\right)}-3.367\times {10}^{-19}\text{J}\\ & =& 1.01\times {10}^{-18}\text{J}-3.367\times {10}^{-19}\text{J}\\ & =& 6.734\times {10}^{-19}\text{J}\\ & & \end{array}$

As $KE=\frac{1}{2}m{v}^2$

Therefore,

$6.734\times {10}^{-19}\text{J}=\frac{1}{2}m{v}^2$………………..(1)

Speed can be calculated using equation (1), such that

$v=\sqrt{\frac{2\times 6.734\times {10}^{-19}}{m}}{\text{m}}^{-1}$

As the mass of an electron $=9.109\times {10}^{-31}\text{kg}$

$\begin{array}{rcl}v& =& \sqrt{\frac{2\times 6.734\times {10}^{-19}}{9.109\times {10}^{-31}}}\text{m}·{\text{s}}^{-1}\\ & =& 1.216\times {10}^6\text{m}·{\text{s}}^{-1}\\ & & \end{array}$