Given that
Metals work function
Na: 2.75 eV, K : 2.30 eV, M_o : 4.17 eV, N_i : 5.15 eV
Wavelength \lambda = 3300 A^{^o} = 3300 * 10^{-10} m
Distance between the laser and photocell = 1 m

Step 1: Calculating the photoelectric emission of the radiation
Energy of proton E=h\upsilon\ =\ \frac{hc}{\lambda}
Where, h - Planck's constant([math] 6.63), speed of light c = 3*10^{8} m/s and \upsilon - frequency.
E = \frac{6.63 * 10^{-34} * 3 * 10^8}{3300 * 10^{-10} }
E = 6 * 10^{-19} joules
E = 3.76 eV \because 1 eV = 1.6* 10^{-19} J
Hence, photoelectric emission of the radiation E = 3.76 eV
Step 2: Metals under photoelectric emission Identification
The photon energy can be passed to the electron if the photon energy is at least as large as the work function.
Na and K:
Work function \phi < E photon energy.
So, photo electric emission is possible
M_o and N_i :
Work function \phi>E photon energy.
So, photo electric emission is not possible
Step 3: Intensity of radiation effect
The intensity of the radiation increases since the laser bought closer to the photocell (50 cm). But photon
energy does not depend on the intensity of the light.
Thus, the energy of incident radiation remains unchanged with a decrease in distance.