The work function for a certain metal is $4.2 \; eV$. Will this metal give photoelectric emission for incident radiation of wavelength $330 \; nm$?

(NO) The work function $\phi_{0} = 4.2 \; eV$.
To find the threshold frequency $\nu_{0}$,we use $\phi_{0} = h\nu_{0}$,where $h = 6.63 \times 10^{-34} \; J \cdot s$.
$\nu_{0} = \frac{\phi_{0}}{h} = \frac{4.2 \times 1.6 \times 10^{-19} \; J}{6.63 \times 10^{-34} \; J \cdot s} \approx 1.01 \times 10^{15} \; Hz$.
The frequency of the incident radiation $\nu$ is given by $\nu = \frac{c}{\lambda}$.
$\nu = \frac{3 \times 10^{8} \; m/s}{330 \times 10^{-9} \; m} \approx 0.91 \times 10^{15} \; Hz$.
Since the incident frequency $\nu \approx 0.91 \times 10^{15} \; Hz$ is less than the threshold frequency $\nu_{0} \approx 1.01 \times 10^{15} \; Hz$,photoelectric emission will not occur.