The metal mainly used in devising photoelectric cells is :

Solution

Chemistry ma’am · 3 steps

A photoelectric cell needs the metal that gives up an electron most easily, and in this group that is caesium.

1. What the cell needs. In a photoelectric cell, light falls on a metal surface and knocks out electrons, which make the current. The metal works best when ordinary visible light is enough to free an electron. So it needs a very low ionisation enthalpy, or work function.
2. Trend down the group. Going down Group 1, atomic size increases and the outer ns¹ electron sits farther from the nucleus. It is held less tightly, so ionisation enthalpy falls from Li to Cs.

   Metal	First ionisation enthalpy (kJ mol⁻¹)	Loses its electron
   Li	520	least easily
   Na	496	less easily
   Rb	403	easily
   Cs	376	most easily

3. So. Caesium has the lowest ionisation enthalpy, so even visible light ejects electrons from it. That is why Cs, option (d), is the metal mainly used in photoelectric cells.

The idea to keep. Whenever a use depends on losing electrons to light, pick the alkali metal with the lowest ionisation enthalpy. Down Group 1 that is caesium, the largest atom.

Answer from the NTA final answer key. Solution by PrepWiser.