The threshold wavelength for ejection of electrons from a metal is 350 nm. The work function for the photoelectric emission from the metal is

Options

A. 8.6×`10^(−19)`J
B. 5.6×`10^(−19)`J
C. 6.6×`10^(−19)`J
D. 5.6×`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 1: Understand the relationship between work function and wavelength
The work function (W) is related to the threshold frequency (ν₀) by the equation:
\[ W = h \nu_0 \]
where \( h \) is Planck's constant.

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.

### Step 2: Substitute the expression for frequency into the work function equation
Substituting the expression for frequency into the work function equation gives:
\[ W = h \left(\frac{c}{\lambda_0}\right) \]

### Step 3: Plug in the 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 \( \lambda_0 = 350 \, \text{nm} = 350 \times 10^{-9} \, \text{m} \)

Now substituting these values into the equation:
\[ W = 6.626 \times 10^{-34} \, \text{J s} \times \left(\frac{3.00 \times 10^8 \, \text{m/s}}{350 \times 10^{-9} \, \text{m}}\right) \]

### Step 4: Calculate the work function
First, calculate the frequency:
\[ \nu_0 = \frac{3.00 \times 10^8}{350 \times 10^{-9}} \]
\[ \nu_0 = \frac{3.00 \times 10^8}{3.50 \times 10^{-7}} \]
\[ \nu_0 \approx 8.57 \times 10^{14} \, \text{Hz} \]

Now calculate the work function:
\[ W = 6.626 \times 10^{-34} \times 8.57 \times 10^{14} \]
\[ W \approx 5.67 \times 10^{-19} \, \text{J} \]

### Final Answer
The work function for the photoelectric emission from the metal is approximately:
\[ W \approx 5.67 \times 10^{-19} \, \text{J} \]