Wavelength is given as $\lambda=3300 \AA$

Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

Planck’s constant $=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}$

Energy of the photon of the incident light can be calculated as,

$E=h c / \lambda=\left(6.63 \times 10^{-34} \times 3 \times 10^{8}\right) / 3300 \times 10^{-10}$

$\Rightarrow 6.018 \times 10^{-19} \mathrm{~J}$

$\Rightarrow\left(6.018 \times 10^{-19} \mathrm{~J}\right) / 1.6 \times 10^{-19}$

$=3.7 \mathrm{eV}$

The incident radiation has a higher energy than the work functions of $Na$ and $K$. For $Mo$ and $Ni$, it is lower. As a result, the photoelectric effect will be absent in $Mo$ and $N$.

The strength of the radiation will grow if the laser is brought closer and positioned $50cm$ away. The radiation’s energy will be unaffected. As a result, the outcome will be the same. The photoelectrons from $Na$ and $K$, on the other hand, will grow in proportion to the intensity.