If the two metals $$A$$ and $$B$$ are exposed to radiation of wavelength 350 nm. The work functions of metals $$A$$ and $$B$$ are 4.8 eV and 2.2 eV. Then choose the correct option

Two metals A and B are exposed to radiation of wavelength $$\lambda = 350$$ nm. Work functions are $$\phi_A = 4.8$$ eV and $$\phi_B = 2.2$$ eV.

$$ E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{350 \text{ nm}} = 3.54 \text{ eV} $$

For photoelectron emission, the photon energy must exceed the work function: $$E \geq \phi$$.

For metal A: $$E = 3.54$$ eV $$< \phi_A = 4.8$$ eV → Metal A will NOT emit photoelectrons.

For metal B: $$E = 3.54$$ eV $$> \phi_B = 2.2$$ eV → Metal B WILL emit photoelectrons.

Only metal A fails to emit photoelectrons, while metal B successfully emits them.

The correct answer is Option D: Metal A will not emit photo-electrons.