When a light of wavelength $\lambda$ falls on the emitter of a photocell, maximum speed of emitted photoelectrons is V. If the incident wavelength is changed to $2 \lambda / 3$, maximum speed of emitted photoelectrons will be

The maximum kinetic energy of photoelectrons in the photoelectric effect is given by $K_{\max} = \frac{1}{2}mV^2 = \frac{hc}{\lambda} - \phi$, where $h$ is Planck’s constant, $c$ is the speed of light, $\lambda$ is the wavelength of incident light, and $\phi$ is the work function of the material.

For incident wavelength $\lambda$ producing maximum speed $V$:
$\frac{1}{2}mV^2 = \frac{hc}{\lambda} - \phi$

For incident wavelength $\frac{2\lambda}{3}$ producing maximum speed $V'$:
$\frac{1}{2}m(V')^2 = \frac{hc}{\frac{2\lambda}{3}} - \phi = \frac{3hc}{2\lambda} - \phi$

Assuming negligible work function ($\phi = 0$) to focus on the relationship:
$\frac{1}{2}mV^2 = \frac{hc}{\lambda}$
$\frac{1}{2}m(V')^2 = \frac{3hc}{2\lambda}$

Dividing these equations:
$\frac{(V')^2}{V^2} = \frac{\frac{3hc}{2\lambda}}{\frac{hc}{\lambda}} = \frac{3}{2}$

Solving for $V'$:
$V' = \sqrt{\frac{3}{2}}V$

The maximum speed of emitted photoelectrons will be $\sqrt{\frac{3}{2}}V$.

Final answer: $\boxed{\text{D}}$