If threshold wavelength `(lambda_(0))` for ejection of electron from metal is 330 nm, then work function for the photoelectron emission is

Options

A. `1.2 xx 10^(-18)J`
B. `1.2 xx 10^(-20)J`
C. `6 xx 10^(-19)J`
D. `6 xx 10^(-12)J`

Explanation

To find the work function for the photoelectron emission given the threshold wavelength, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data**:
- Threshold wavelength, \( \lambda_0 = 330 \, \text{nm} \)

2. **Convert the wavelength from nanometers to meters**:
\[ \lambda_0 = 330 \, \text{nm} = 330 \times 10^{-9} \, \text{m} \]

3. **Use the formula for work function**:
The work function \( W_0 \) can be calculated using the equation:
\[ W_0 = \frac{hc}{\lambda_0} \]
where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

4. **Substitute the values into the equation**:
\[ W_0 = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{330 \times 10^{-9} \, \text{m}} \]

5. **Calculate the numerator**:
\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Jm} \]

6. **Calculate the work function**:
\[ W_0 = \frac{1.9878 \times 10^{-25}}{330 \times 10^{-9}} = \frac{1.9878 \times 10^{-25}}{3.3 \times 10^{-7}} = 6.02 \times 10^{-19} \, \text{J} \]

7. **Final answer**:
The work function for the photoelectron emission is approximately:
\[ W_0 \approx 6.02 \times 10^{-19} \, \text{J} \]