The threshold wavelength for ejection of electrons from a metal is 280 nm. The work function for the photoelectric emission from the metal is

Options

A. 7×`10^(−19)`J
B. 5.6×`10^(−19)`J
C. 3.6×`10^(−19)`J
D. 4.6×`10^(−19)`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. **Understand the Relationship**: The work function (W) in the photoelectric effect can be expressed in terms of the threshold frequency (ν₀) as:
\[ W = h \nu_0 \]
where \( h \) is Planck's constant.

2. **Convert Threshold Wavelength to Frequency**: The threshold frequency can also be expressed in terms of the threshold wavelength (λ₀):
\[ \nu_0 = \frac{c}{\lambda_0} \]
where \( c \) is the speed of light.

3. **Substituting the Expression**: Substituting the expression for frequency into the work function equation gives:
\[ W = h \frac{c}{\lambda_0} \]

4. **Insert Known Values**: We know:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- Speed of light \( c = 3.00 \times 10^8 \, \text{m/s} \)
- Threshold wavelength \( \lambda_0 = 280 \, \text{nm} = 280 \times 10^{-9} \, \text{m} \)

5. **Calculate Work Function**:
\[ W = h \frac{c}{\lambda_0} = (6.626 \times 10^{-34} \, \text{J s}) \cdot \left(\frac{3.00 \times 10^8 \, \text{m/s}}{280 \times 10^{-9} \, \text{m}}\right) \]

6. **Perform the Calculation**:
- First, calculate \( \frac{c}{\lambda_0} \):
\[ \frac{3.00 \times 10^8}{280 \times 10^{-9}} = 1.0714 \times 10^{15} \, \text{Hz} \]
- Now, calculate \( W \):
\[ W = 6.626 \times 10^{-34} \cdot 1.0714 \times 10^{15} \approx 7.1 \times 10^{-19} \, \text{J} \]

7. **Final Result**: The work function for the photoelectric emission from the metal is approximately:
\[ W \approx 7.1 \times 10^{-19} \, \text{J} \]