The threshold wavelength for ejection of electrons from a metal is 300 nm. The work function for the photoelectric emission from the metal is

Options

A. 6.6×`10^(−19)`J
B. 5.6×`10^(−19)`J
C. 7.6×`10^(−19)`J
D. 5.6×`10^(−21)`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 Information**:
- Threshold wavelength (\( \lambda_0 \)) = 300 nm

2. **Convert Wavelength to Meters**:
- Since 1 nm = \( 10^{-9} \) m, we convert 300 nm to meters:
\[ \lambda_0 = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m} = 3 \times 10^{-7} \, \text{m} \]

3. **Use the Work Function Formula**:
- The work function (\( W \)) is given by the formula:
\[ W = \frac{hc}{\lambda_0} \]
- 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**:
- Plugging in the values:
\[ W = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^{8} \, \text{m/s})}{3 \times 10^{-7} \, \text{m}} \]

5. **Calculate the Work Function**:
- First, calculate the numerator:
\[ hc = (6.626 \times 10^{-34})(3.00 \times 10^{8}) = 1.9878 \times 10^{-25} \, \text{J m} \]
- Now, divide by the wavelength:
\[ W = \frac{1.9878 \times 10^{-25} \, \text{J m}}{3 \times 10^{-7} \, \text{m}} = 6.626 \times 10^{-19} \, \text{J} \]

6. **Final Result**:
- The work function for the photoelectric emission from the metal is approximately:
\[ W \approx 6.626 \times 10^{-19} \, \text{J} \]

### Conclusion:
The work function for the photoelectric emission from the metal is \( 6.626 \times 10^{-19} \, \text{J} \).

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