For a metal of work function 6.6 eV, which of the following...

The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \dfrac{hc}{\Phi}$.

Substituting the given values:

$\lambda_0 = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 1.6 \times 10^{-19}}$

$\lambda_0 = \dfrac{3}{1.6} \times 10^{-7}$ m

$\lambda_0 = 1.875 \times 10^{-7}$ m $= 187.5$ nm

For the photoelectric effect to take place, the wavelength of incident radiation must be less than or equal to the threshold wavelength ($\lambda \le \lambda_0$).

Among the given options, $200$ nm is greater than $187.5$ nm.

Thus, radiation of wavelength $200$ nm will not give rise to the photoelectric effect.

Answer: $200$ nm