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

Options

A. `1.2xx10^(-18)J`
B. `1.2xx10^(-20)J`
C. `6xx10^(-19)J`
D. `6xx10^(-12)J`

Explanation

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

### Step-by-Step Solution:

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

2. **Convert the wavelength from nanometers to meters:**
- \( 1 \, \text{nm} = 10^{-9} \, \text{m} \)
- Therefore, \( \lambda_0 = 330 \, \text{nm} = 330 \times 10^{-9} \, \text{m} \)

3. **Use the formula for work function:**
- The work function \( \phi \) is given by the formula:
\[ \phi = \frac{hc}{\lambda_0} \]
- Where:
- \( h \) (Planck's constant) \( = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) (speed of light) \( = 3.00 \times 10^8 \, \text{m/s} \)

4. **Substitute the values into the formula:**
\[ \phi = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^8 \, \text{m/s})}{330 \times 10^{-9} \, \text{m}} \]

5. **Calculate the numerator:**
\[ 6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25} \, \text{J m} \]

6. **Calculate the work function:**
\[ \phi = \frac{1.9878 \times 10^{-25} \, \text{J m}}{330 \times 10^{-9} \, \text{m}} = \frac{1.9878 \times 10^{-25}}{3.30 \times 10^{-7}} \approx 6.01 \times 10^{-19} \, \text{J} \]

7. **Final result:**
- The work function \( \phi \approx 6.01 \times 10^{-19} \, \text{J} \)

### Conclusion:
The work function for photoelectric emission is approximately \( 6.01 \times 10^{-19} \, \text{J} \).