Work functions of four metals M$_1$, M$_2$, M$_3$ and M$_4$ are 4.8, 4.3, 4.75 and 3.75 eV respectively. The metals which do not show photoelectric effect when light of wavelength 310 nm falls on the metals are
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The condition for the photoelectric effect is that the incident photon's energy must be greater than the metal's work function ($E_{photon} \ge \phi$). Use the formula $E(\text{eV}) = 1240/\lambda(\text{nm})$ for quick energy calculations.
We are asked to determine which metals do not show the photoelectric effect when illuminated with light of wavelength $\lambda = 310$ nm. Step 1: Recall the condition for photoelectric effect
The photoelectric effect occurs only if the energy of the incident photon ($E_\text{photon}$) is greater than or equal to the work function ($\phi$) of the metal:
\[
E_\text{photon} \ge \phi
\]
If $E_\text{photon}<\phi$, the metal does not exhibit the photoelectric effect.
Step 2: Calculate the photon energy
The energy of a photon is given by:
\[
E_\text{photon} = \frac{hc}{\lambda} \approx \frac{1240}{\lambda \, (\text{nm})} \, \text{eV}
\]
Given $\lambda = 310$ nm:
\[
E_\text{photon} = \frac{1240}{310} \approx 4.0 \, \text{eV}
\]
Step 3: Compare photon energy with work functions of the metals
M$_1$: $\phi_1 = 4.8$ eV $\Rightarrow 4.8>4.0$ ��� No photoelectric effect
M$_2$: $\phi_2 = 4.3$ eV $\Rightarrow 4.3>4.0$ ��� No photoelectric effect
M$_3$: $\phi_3 = 4.75$ eV $\Rightarrow 4.75>4.0$ ��� No photoelectric effect
Step 4: Identify metals that do not show the effect
Based on the above, metals that do not show the photoelectric effect are:
\[
\text{M}_1, \text{M}_2, \text{M}_3
\]
Note:
The answer key states (B) M$_1$, M$_3$ only. This appears to be inconsistent with the given work function values and wavelength. The correct metals not showing the effect based on the numbers provided are M$_1$, M$_2$, M$_3$ (Option C).
