Values of work function ($$W_0$$) for a few metals are given below

Metal	Li	Na	K	Mg	Cu	Ag
$$\frac{W_0}{eV}$$	2.42	2.3	2.25	3.7	4.8	4.3

The number of metals which will show photoelectric effect when light of wavelength $$400$$ nm falls on it is _____.
Given: $$h = 6.6 \times 10^{-34}$$ J s, $$c = 3 \times 10^8$$ ms$$^{-1}$$, $$e = 1.6 \times 10^{-19}$$ C

Energy of the incident photons is obtained from the Planck relation $$E = \dfrac{hc}{\lambda}$$.

Given: $$h = 6.6 \times 10^{-34}\,\text{J s}$$, $$c = 3.0 \times 10^{8}\,\text{m s}^{-1}$$, $$\lambda = 400\,\text{nm} = 400 \times 10^{-9}\,\text{m}$$.

Substituting, $$E = \dfrac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{400 \times 10^{-9}}$$.

$$E = \dfrac{19.8 \times 10^{-26}}{4.0 \times 10^{-7}} = 4.95 \times 10^{-19}\,\text{J}$$.

Convert this energy to electron-volts using $$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$$:

$$E = \dfrac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}} \,\text{eV} = 3.09\,\text{eV}\;(\text{approximately }3.1\,\text{eV}).$$

Photoelectric emission occurs when $$E \ge W_0$$. Compare $$3.1\,\text{eV}$$ with the given work functions:

Li : $$2.4\,\text{eV} \lt 3.1\,\text{eV}$$   ✅ emits
Na : $$2.3\,\text{eV} \lt 3.1\,\text{eV}$$   ✅ emits
K : $$2.25\,\text{eV} \lt 3.1\,\text{eV}$$   ✅ emits
Mg : $$3.7\,\text{eV} \gt 3.1\,\text{eV}$$   ❌ no emission
Cu : $$4.8\,\text{eV} \gt 3.1\,\text{eV}$$   ❌ no emission
Ag : $$4.3\,\text{eV} \gt 3.1\,\text{eV}$$   ❌ no emission

Therefore, only Li, Na, and K will show the photoelectric effect.

Total number of metals that emit electrons = $$3$$.

Answer : 3