Use the photoelectric condition

A metal emits photoelectrons only if the incident photon energy is at least equal to its work function: $E_{\gamma} \ge \phi$

where $E_{\gamma} = \frac{hc}{\lambda}$

Calculate photon energy for $\lambda = 350\,\text{nm}$

Using the standard relation in electron-volts: $E_{\gamma}(\text{eV}) = \frac{1240}{\lambda(\text{nm})}$

So, $E_{\gamma} = \frac{1240}{350} \approx 3.54\,\text{eV}$

Compare with work functions

• For metal A: $\phi_A = 4.8\,\text{eV}$ Since $3.54 < 4.8$ metal A will not emit photoelectrons.

• For metal B: $\phi_B = 2.2\,\text{eV}$ Since $3.54 > 2.2$ metal B will emit photoelectrons.

Check options

• A: Metal B will not emit photo-electrons → False
• B: Both metals A and B will not emit photo-electrons → False
• C: Metal A will not emit photo-electrons → True
• D: Both metals A and B will emit photo-electrons → False

Final answer

The correct option is: $\boxed{\text{C}}$