The threshold wavelength for ejection of electrons from a metal is 150 nm. The work function for the photoelectric emission from the metal is

Options

A. 5.6×`10^(−21)`J
B. 1.3×`10^(−18)`J
C. 6.6×`10^(−19)`J
D. 4.6×`10^(−21)`J

Explanation

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

### Step 1: Understand the relationship between work function and threshold wavelength
The work function (W) can be expressed in terms of the threshold frequency (ν₀) using the equation:
\[ W = h \nu_0 \]
where \( h \) is Planck's constant.

### Step 2: Relate threshold frequency to threshold wavelength
The threshold frequency can also be related to the threshold wavelength (λ₀) using the equation:
\[ \nu_0 = \frac{c}{\lambda_0} \]
where \( c \) is the speed of light. Therefore, we can express the work function in terms of the threshold wavelength:
\[ W = h \frac{c}{\lambda_0} \]

### Step 3: Substitute the known values
Given that the threshold wavelength \( \lambda_0 \) is 150 nm, we first convert this to meters:
\[ \lambda_0 = 150 \, \text{nm} = 150 \times 10^{-9} \, \text{m} \]

Now, we can substitute the values into the equation:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- Speed of light \( c = 3.00 \times 10^{8} \, \text{m/s} \)

Thus, we have:
\[ W = h \frac{c}{\lambda_0} = 6.626 \times 10^{-34} \times \frac{3.00 \times 10^{8}}{150 \times 10^{-9}} \]

### Step 4: Calculate the work function
Now we perform the calculation:
1. Calculate \( \frac{c}{\lambda_0} \):
\[ \frac{3.00 \times 10^{8}}{150 \times 10^{-9}} = 2.00 \times 10^{15} \, \text{Hz} \]

2. Now multiply by \( h \):
\[ W = 6.626 \times 10^{-34} \times 2.00 \times 10^{15} \]
\[ W = 1.3252 \times 10^{-18} \, \text{J} \]

### Step 5: Round the answer
Rounding the answer gives:
\[ W \approx 1.32 \times 10^{-18} \, \text{J} \]

### Conclusion
The work function for the photoelectric emission from the metal is approximately:
\[ \boxed{1.32 \times 10^{-18} \, \text{J}} \]

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