A metal has a threshold wavelength fo `6000 Å`. Calculate (i) threshold frequency (ii) work function. Will there be photoelectric emission or not ?

Answer

To solve the problem step by step, we will calculate the threshold frequency and work function for a metal with a threshold wavelength of 6000 Å.

### Step 1: Convert the threshold wavelength from Angstroms to meters.
The threshold wavelength (λ) is given as 6000 Å. We know that:
1 Å = \(10^{-10}\) meters.

So,
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6.0 \times 10^{-7} \, \text{m}. \]

### Step 2: Calculate the threshold frequency (ν₀).
We use the formula that relates wavelength and frequency:
\[ c = \lambda \cdot \nu, \]
where \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)).

Rearranging gives:
\[ \nu = \frac{c}{\lambda}. \]

Substituting the values:
\[ \nu_0 = \frac{3 \times 10^8 \, \text{m/s}}{6.0 \times 10^{-7} \, \text{m}}. \]

Calculating this:
\[ \nu_0 = 0.5 \times 10^{15} \, \text{Hz} = 5.0 \times 10^{14} \, \text{Hz}. \]

### Step 3: Calculate the work function (W).
The work function can be calculated using the formula:
\[ W = h \cdot \nu_0, \]
where \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34} \, \text{J s}\).

Substituting the values:
\[ W = 6.626 \times 10^{-34} \, \text{J s} \cdot 5.0 \times 10^{14} \, \text{Hz}. \]

Calculating this:
\[ W = 3.313 \times 10^{-19} \, \text{J}. \]

### Step 4: Convert the work function from Joules to electron volts (eV).
To convert Joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ W \, (\text{eV}) = \frac{3.313 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}}. \]

Calculating this:
\[ W \, (\text{eV}) = 2.07 \, \text{eV} \, (\text{approximately}). \]

### Step 5: Determine if photoelectric emission will occur.
Photoelectric emission will occur if the energy of the incident photons is greater than the work function. The energy of a photon can be calculated using:
\[ E = h \cdot \nu, \]
where \(\nu\) is the frequency of the incident light. Since we have calculated the threshold frequency, any frequency above \(5.0 \times 10^{14} \, \text{Hz}\) will result in photoelectric emission.

### Summary of Results:
1. **Threshold Frequency (ν₀)**: \(5.0 \times 10^{14} \, \text{Hz}\)
2. **Work Function (W)**: \(2.07 \, \text{eV}\)
3. **Photoelectric Emission**: Yes, if the frequency of incident light is greater than \(5.0 \times 10^{14} \, \text{Hz}\).