Step 1: Understand the photoelectric condition.
A metal emits electrons only when the light photon carries enough energy. The needed minimum is the work function $\phi$. The condition is: emission happens when photon energy $E \ge \phi$. If the photon energy is smaller than $\phi$, no electrons come out.
Step 2: Recall the easy energy formula.
The energy of one photon in electron volts is found quickly from its wavelength using \[ E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}. \] This shortcut comes from $E = \dfrac{hc}{\lambda}$ with the constants already combined.
Step 3: Find the photon energy.
Here the wavelength is $300$ nm, so \[ E = \frac{1240}{300} \approx 4.13 \ \text{eV}. \] Every photon in this light carries about $4.13$ eV.
Step 4: Compare with Mg and Cu.
For Mg, $\phi = 3.7$ eV. Since $4.13 > 3.7$, Mg emits electrons. For Cu, $\phi = 4.8$ eV. Since $4.13 < 4.8$, Cu does not emit.
Step 5: Compare with Ag and Li.
For Ag, $\phi = 4.3$ eV. Since $4.13 < 4.3$, Ag does not emit. For Li, $\phi = 2.5$ eV. Since $4.13 > 2.5$, Li emits electrons.
Step 6: Collect the answer.
Only the metals with work function below $4.13$ eV show the effect. These are Mg and Li. \[ \boxed{\text{Mg and Li}} \]