The threshold wavelength for photoelectric effect of a metal is 6500 Å. The work function of the metal is approximately

Options

A. 2 eV
B. 1 eV
C. 0.1 eV
D. 3 eV

Answer

To find the work function of a metal given its threshold wavelength for the photoelectric effect, we can use the formula:

\[ \phi = \frac{hc}{\lambda_0} \]

Where:
- \(\phi\) is the work function (in electron volts, eV),
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda_0\) is the threshold wavelength (in meters).

### Step 1: Convert the threshold wavelength from angstroms to meters.
Given:
\[ \lambda_0 = 6500 \, \text{Å} = 6500 \times 10^{-10} \, \text{m} = 6.5 \times 10^{-7} \, \text{m} \]

### Step 2: Calculate \(hc\).
Using the values of \(h\) and \(c\):
\[ hc = (6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s}) = 1.9878 \times 10^{-25} \, \text{Jm} \]

### Step 3: Calculate the work function \(\phi\).
Substituting the values into the work function formula:
\[ \phi = \frac{hc}{\lambda_0} = \frac{1.9878 \times 10^{-25} \, \text{Jm}}{6.5 \times 10^{-7} \, \text{m}} = 3.055 \times 10^{-19} \, \text{J} \]

### Step 4: Convert the work function from joules to electron volts.
Using the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ \phi = \frac{3.055 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 1.907 \, \text{eV} \]

### Conclusion:
The work function of the metal is approximately **1.9 eV**.