To determine the number of metals that will show the photoelectric effect when light of 300 nm wavelength falls on them, we first need to calculate the energy (in eV) of the incident photons. The energy ($E$) of a photon can be calculated using the formula:

$E = \frac{hc}{\lambda}$

where

• $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s),


• $c$ is the speed of light ($3.00 \times 10^{8}$ m/s),


• $\lambda$ is the wavelength of the light (in meters).

First, convert the wavelength from nm to meters:

$300\, \text{nm} = 300 \times 10^{-9}\, \text{m}$

Then, calculate the energy of the photons:

$E = \frac{(6.626 \times 10^{-34}\, \text{J·s}) \times (3.00 \times 10^{8}\, \text{m/s})}{300 \times 10^{-9}\, \text{m}}$

$E = \frac{(6.626 \times 3.00) \times 10^{-19}}{300}\, \text{J}$

$E = \frac{19.878 \times 10^{-19}}{300}\, \text{J}$

$E \approx 6.626 \times 10^{-19}\, \text{J}$

To convert the energy from joules to electron volts (eV), we divide by the charge of an electron ($1.602 \times 10^{-19}$ C):

$E \approx \frac{6.626 \times 10^{-19}\, \text{J}}{1.602 \times 10^{-19}\, \text{C/e^-}} \approx 4.14\, \text{eV}$

Now, to determine whether the photoelectric effect will occur, we compare the energy of the incident photons to the work function ($\phi$) of each metal. The photoelectric effect occurs if the photon energy is greater than or equal to the work function of the metal:

• For Li ($\phi = 2.4\, \text{eV}$): Yes, $4.14\, \text{eV} > 2.4\, \text{eV}$


• For Na ($\phi = 2.3\, \text{eV}$): Yes, $4.14\, \text{eV} > 2.3\, \text{eV}$


• For K ($\phi = 2.2\, \text{eV}$): Yes, $4.14\, \text{eV} > 2.2\, \text{eV}$


• For Mg ($\phi = 3.7\, \text{eV}$): Yes, $4.14\, \text{eV} > 3.7\, \text{eV}$


• For Cu ($\phi = 4.8\, \text{eV}$): No, $4.14\, \text{eV} < 4.8\, \text{eV}$


• For Ag ($\phi = 4.3\, \text{eV}$): No, $4.14\, \text{eV} < 4.3\, \text{eV}$


• For Fe ($\phi = 4.7\, \text{eV}$): No, $4.14\, \text{eV} < 4.7\, \text{eV}$


• For Pt ($\phi = 6.3\, \text{eV}$): No, $4.14\, \text{eV} < 6.3\, \text{eV}$


• For W ($\phi = 4.75\, \text{eV}$): No, $4.14\, \text{eV} < 4.75\, \text{eV}$

Thus, the number of metals which will show the photoelectric effect when light of 300 nm wavelength falls on the metal is 4 (Li, Na, K, Mg).