To determine which metals will show the photoelectric effect when exposed to light of 300 nm wavelength, we need to follow these steps:
### Step-by-Step Solution:
1. **Understand the Photoelectric Effect**:
The photoelectric effect occurs when light of sufficient energy strikes a metal surface, causing the ejection of electrons. The energy of the incident light must be greater than the work function (Φ) of the metal.
2. **Calculate the Energy of the Incident Light**:
The energy (E) of the incident light can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \(h\) (Planck's constant) = \(6.626 \times 10^{-34} \, \text{J s}\)
- \(c\) (speed of light) = \(3 \times 10^{8} \, \text{m/s}\)
- \(\lambda\) (wavelength) = \(300 \, \text{nm} = 300 \times 10^{-9} \, \text{m}\)
Substituting the values:
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^{8} \, \text{m/s})}{300 \times 10^{-9} \, \text{m}}
\]
3. **Perform the Calculation**:
\[
E = \frac{(6.626 \times 3) \times 10^{-34 + 8}}{300 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{300 \times 10^{-9}} = \frac{19.878}{300} \times 10^{-17} \, \text{J}
\]
\[
E \approx 6.626 \times 10^{-19} \, \text{J}
\]
4. **Convert Energy from Joules to Electron Volts**:
To convert the energy from Joules to electron volts (eV), use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[
E \, (\text{in eV}) = \frac{6.626 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 4.14 \, \text{eV}
\]
5. **Compare with Work Functions of Metals**:
Now, we compare the calculated energy (4.14 eV) with the work functions of the given metals:
- Li: 2.4 eV
- Na: 2.3 eV
- K: 2.2 eV
- Mg: 3.7 eV
- Cu: 4.8 eV
- Ag: 4.3 eV
- Fe: 4.7 eV
- Pt: 6.3 eV
- W: 4.75 eV
The metals that will show the photoelectric effect are those with a work function less than 4.14 eV:
- Li (2.4 eV) < 4.14 eV
- Na (2.3 eV) < 4.14 eV
- K (2.2 eV) < 4.14 eV
- Mg (3.7 eV) < 4.14 eV
Metals with work functions greater than 4.14 eV (Cu, Ag, Fe, Pt, W) will not show the photoelectric effect.
### Final Answer:
The metals that will show the photoelectric effect when light of 300 nm wavelength falls on them are **Lithium (Li), Sodium (Na), Potassium (K), and Magnesium (Mg)**.
