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 the photoelectric emission given the threshold wavelength, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The work function (ϕ) is related to the threshold wavelength (λ₀) by the equation:
\[ \phi = \frac{hc}{\lambda_0} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda_0 \) is the threshold wavelength.

2. **Convert Wavelength to Meters**: The given threshold wavelength is 330 nm. We need to convert this into meters:
\[ \lambda_0 = 330 \, \text{nm} = 330 \times 10^{-9} \, \text{m} \]

3. **Substitute Constants**: Use the known values for Planck's constant and the speed of light:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3.00 \times 10^8 \, \text{m/s} \)

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

5. **Perform the Calculation**:
- First calculate \( hc \):
\[ hc = 6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25} \, \text{J m} \]
- Now divide by \( \lambda_0 \):
\[ \phi = \frac{1.9878 \times 10^{-25}}{330 \times 10^{-9}} \approx 6.02 \times 10^{-19} \, \text{J} \]

6. **Final Result**: The work function for the photoelectric emission is approximately:
\[ \phi \approx 6.02 \times 10^{-19} \, \text{J} \]