The photoelectric effect occurs when a photon with energy greater than the work function of a metal strikes the metal surface, causing electrons to be emitted. The minimum energy required for this process is given by the equation:

$\begin{equation} E = h\nu = \frac{hc}{\lambda} \end{equation}$

where $E$ is the energy of the photon, $h$ is Planck's constant, $\nu$ is the frequency of the photon, $c$ is the speed of light, $\lambda$ is the wavelength of the photon.

If the energy of the photon is greater than the work function ($W_0$) of the metal, then electrons will be emitted. The energy of the photon is related to its wavelength by the above equation.

For the given wavelength of 400 nm, the energy of the photon is:

$\begin{equation} E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \text{ J s} \times 3.0 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} = 4.97 \times 10^{-19} \text{ J} \end{equation}$

For the photoelectric effect to occur, this energy must be greater than the work function of the metal. The metals that will show the photoelectric effect can be determined by comparing the energy of the photon to the work function of each metal.

Metal	$W_0$ (eV)	$W_0$ (J)	$E > W_0$?
Li	2.42	3.88 $\times$ 10$^{-19}$ J	Yes
Na	2.3	3.68 $\times$ 10$^{-19}$ J	Yes
K	2.25	3.60 $\times$ 10$^{-19}$ J	Yes
Mg	3.7	5.92 $\times$ 10$^{-19}$ J	No
Cu	4.8	7.68 $\times$ 10$^{-19}$ J	No
Ag	4.3	6.88 $\times$ 10$^{-19}$ J	No

From the table, we see that three metals (Li, Na, and K) have work functions less than the energy of the photon with a wavelength of 400 nm. Therefore, these three metals will show the photoelectric effect when light of this wavelength falls on them.

Thus, the number of metals which will show photoelectric effect when light of wavelength 400 nm falls on them is 3.