The energy of the incident photon is: \[ E = \frac{1240}{\lambda} = \frac{1240}{550} \approx 2.25 \, \text{eV} \] For the photoelectric effect to occur, the energy of the photon must be greater than the work function of the metal.
For Cs (work function = 1.9 eV):
Since \( 2.25 \, \text{eV} > 1.9 \, \text{eV} \), photoelectric effect is possible for Cs.
For Li (work function = 2.5 eV):
Since \( 2.25 \, \text{eV} <2.5 \, \text{eV} \), photoelectric effect is not possible for Li.
Thus, the answer is \( \boxed{\text{Cs only}} \).
Which of the following statements are correct, if the threshold frequency of caesium is $ 5.16 \times 10^{14} \, \text{Hz} $?
The net current flowing in the given circuit is ___ A.
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .