To determine which metals do not show the photoelectric effect when light of wavelength 310 nm falls on them, we need to compare the energy of the incident light with the work functions of the metals.
The energy of a photon (\(E\)) can be calculated using the formula:
\(E = \frac{hc}{\lambda}\)
Where:
\(h\) = Planck's constant = \(6.626 \times 10^{-34}\) Js
\(c\) = Speed of light = \(3 \times 10^8\) m/s
\(\lambda\) = Wavelength of the light in meters.
Given, \(\lambda = 310 \, \text{nm} = 310 \times 10^{-9} \, \text{m}\).
Substituting the values, we get:
\(E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{310 \times 10^{-9}}\)
\(E \approx 6.409 \times 10^{-19} \, \text{J}\)
Convert this energy to electron volts (eV) since work functions are given in eV:
\(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\)
\(E \approx \frac{6.409 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.00 \, \text{eV}\)
Comparing the energy of the incident photon (4.00 eV) with the work functions of the metals:
A photoelectric effect occurs if the photon's energy is greater than or equal to the metal's work function. Here, M1 and M3 have work functions greater than the photon's energy, hence they do not show the photoelectric effect.
Therefore, the metals M1 and M3 do not show the photoelectric effect when subjected to light of wavelength 310 nm.
Correct Answer: M1, M3 only