The threshold wavelength for ejection of electrons from a metal is 430 nm. The work function for the photoelectric emission from the metal is

Options

A. 4.6×`10^(−19)`J
B. 6.0×`10^(−21)`J
C. 6.0×`10^(−19)`J
D. 4.6×`10^(−20)`J

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Threshold wavelength (\( \lambda_0 \)) = 430 nm

2. **Convert Wavelength to Meters:**
- Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \), we convert the wavelength:
\[ \lambda_0 = 430 \, \text{nm} = 430 \times 10^{-9} \, \text{m} \]

3. **Use the Formula for Work Function:**
- The work function (\( W \)) is related to the threshold frequency (\( \nu_0 \)) and can be expressed as:
\[ W = h \nu_0 \]
where \( h \) is Planck's constant.

4. **Relate Frequency to Wavelength:**
- The threshold frequency (\( \nu_0 \)) can be calculated using the speed of light (\( c \)) and the wavelength:
\[ \nu_0 = \frac{c}{\lambda_0} \]
where \( c = 3 \times 10^8 \, \text{m/s} \).

5. **Substitute the Values:**
- Now substitute \( \lambda_0 \) into the frequency formula:
\[ \nu_0 = \frac{3 \times 10^8 \, \text{m/s}}{430 \times 10^{-9} \, \text{m}} \]

6. **Calculate the Frequency:**
- Performing the calculation:
\[ \nu_0 \approx \frac{3 \times 10^8}{430 \times 10^{-9}} \approx 6.9767 \times 10^{14} \, \text{Hz} \]

7. **Calculate the Work Function:**
- Now substitute \( \nu_0 \) back into the work function formula:
\[ W = h \nu_0 \]
where \( h = 6.626 \times 10^{-34} \, \text{Js} \):
\[ W = 6.626 \times 10^{-34} \times 6.9767 \times 10^{14} \]

8. **Perform the Final Calculation:**
- Calculating the work function:
\[ W \approx 4.62 \times 10^{-19} \, \text{J} \]

### Final Answer:
The work function for the photoelectric emission from the metal is approximately \( 4.62 \times 10^{-19} \, \text{J} \).

---