The photoelectric work function for a metal surface is `4.125 eV`. The cut - off wavelength for this surface is

Options

A. `4125 Å`
B. `2062.5 Å`
C. `3000 Å`
D. `6000 Å`

Explanation

To find the cut-off wavelength for a metal surface with a given photoelectric work function, we can use the relationship between energy, wavelength, and the work function. Here’s a step-by-step solution:

### Step 1: Understand the relationship between energy and wavelength
The energy of a photon can be expressed in terms of its wavelength using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \, \text{eV s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 2: Convert the work function from eV to joules
The work function (\( \phi \)) is given as \( 4.125 \, \text{eV} \). To use the formula, we can keep it in eV since we will use the energy in eV.

### Step 3: Set up the equation for cut-off wavelength
At the cut-off wavelength, the energy of the incoming photon is equal to the work function:
\[ E = \phi \]
Thus, we can write:
\[ \frac{hc}{\lambda} = \phi \]

### Step 4: Rearrange the equation to solve for wavelength
Rearranging the equation gives us:
\[ \lambda = \frac{hc}{\phi} \]

### Step 5: Substitute the known values
Substituting the values:
- \( h = 4.135667696 \times 10^{-15} \, \text{eV s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( \phi = 4.125 \, \text{eV} \)

We can calculate \( \lambda \):
\[ \lambda = \frac{(4.135667696 \times 10^{-15} \, \text{eV s}) \times (3 \times 10^8 \, \text{m/s})}{4.125 \, \text{eV}} \]

### Step 6: Calculate the wavelength
Calculating this gives:
\[ \lambda = \frac{1.240 \times 10^{-6} \, \text{m}}{4.125} \]
\[ \lambda = 3.00 \times 10^{-7} \, \text{m} \]

### Step 7: Convert meters to angstroms
Since \( 1 \, \text{angstrom} = 10^{-10} \, \text{m} \):
\[ \lambda = 3.00 \times 10^{-7} \, \text{m} \times \frac{10^{10} \, \text{angstrom}}{1 \, \text{m}} = 3000 \, \text{angstroms} \]

### Final Answer
The cut-off wavelength for the metal surface is \( 3000 \, \text{angstroms} \).

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