Photons of what minimum frequency are required to remove electrons from gold?
Note: The work function for gold is 4.8 eV.

Options

A. `7.3 xx 10^(14) Hz`
B. `3.8 xx 10^(17) Hz`
C. `1.2 xx 10^(15) Hz`
D. `6.5xx 10^(14) Hz`

Answer

To find the minimum frequency of photons required to remove electrons from gold, we can use the concept of the photoelectric effect. The minimum frequency, also known as the threshold frequency (ν₀), is related to the work function (φ) of the material. The work function for gold is given as 4.8 eV.

### Step-by-Step Solution:

1. **Understand the relationship between work function and frequency**:
The energy of a photon is given by the equation:
\[ E = h \cdot \nu \]
where \(E\) is the energy of the photon, \(h\) is Planck's constant, and \(\nu\) is the frequency of the photon. The work function (φ) is the minimum energy required to remove an electron from the surface of a material.

2. **Convert work function from eV to Joules**:
The work function for gold is given as 4.8 eV. To convert this to Joules, we use the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\):
\[ \phi = 4.8 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 7.68 \times 10^{-19} \, \text{J} \]

3. **Use the equation to find the threshold frequency**:
Rearranging the equation \(E = h \cdot \nu\) to solve for frequency gives:
\[ \nu = \frac{\phi}{h} \]
Substituting the values for φ and \(h\) (where \(h = 6.63 \times 10^{-34} \, \text{J s}\)):
\[ \nu = \frac{7.68 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J s}} \approx 1.16 \times 10^{15} \, \text{Hz} \]

4. **Conclusion**:
The minimum frequency of photons required to remove electrons from gold is approximately \(1.16 \times 10^{15} \, \text{Hz}\).