In a photoelectric cell, for the incident radiation of wavelength λ...

Using Einstein's photoelectric equation:

$\dfrac{1}{2}mv^2 = \dfrac{hc}{\lambda} - \phi$

When the wavelength is changed to $\dfrac{\lambda}{3}$, the new maximum kinetic energy is:

$\dfrac{1}{2}mv'^2 = \dfrac{3hc}{\lambda} - \phi$

Substituting $\dfrac{hc}{\lambda} = \dfrac{1}{2}mv^2 + \phi$ into the second equation:

$\dfrac{1}{2}mv'^2 = 3\left(\dfrac{1}{2}mv^2 + \phi\right) - \phi$

$\dfrac{1}{2}mv'^2 = 3\left(\dfrac{1}{2}mv^2\right) + 2\phi$

Since the work function $\phi > 0$, we get:

$\dfrac{1}{2}mv'^2 > 3 \left(\dfrac{1}{2}mv^2\right)$

$v'^2 > 3v^2$

$v' > v\sqrt{3}$

Thus, the speed of the fastest emitted electron is greater than $v(\sqrt{3}).$

Answer: greater than $v(\sqrt{3})$