The work function for a metal is 4eV. To emit a photoelectron of zero velocity from the surface of the metal, the wavelength of incident light should be :

Options

A. `2700Å`
B. `1700Å`
C. `5900Å`
D. `3100Å`

Explanation

To solve the problem of finding the wavelength of incident light required to emit a photoelectron of zero velocity from a metal with a work function of 4 eV, we can follow these steps:

### Step 1: Understand the Concept
The work function (Φ) is the minimum energy required to remove an electron from the surface of a metal. When light of a certain wavelength strikes the metal, it can provide energy to the electrons. If the energy provided by the light is equal to the work function, the electron will be emitted with zero kinetic energy (i.e., zero velocity).

### Step 2: Use the Energy-Wavelength Relationship
The energy (E) of a photon can be expressed in terms of its wavelength (λ) using the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the light.

### Step 3: Set the Energy Equal to the Work Function
Since we want the energy of the photon to be equal to the work function (Φ = 4 eV), we can set up the equation:
\[ \frac{hc}{\lambda} = \Phi \]
Substituting the known values:
\[ \frac{hc}{\lambda} = 4 \text{ eV} \]

### Step 4: Convert Planck's Constant to Electron Volts
Planck's constant in joules is:
\[ h = 6.63 \times 10^{-34} \text{ J s} \]
To convert this to electron volts, we can use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ h = \frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} \text{ eV s} \approx 4.14 \times 10^{-15} \text{ eV s} \]

### Step 5: Substitute Values into the Equation
Now substituting \( h \) and \( c \) into the equation:
- Speed of light \( c = 3 \times 10^8 \text{ m/s} \)
- Convert \( c \) to angstroms (1 m = \( 10^{10} \) angstroms):
\[ c = 3 \times 10^{18} \text{ angstrom/s} \]

Now substituting into the equation:
\[ \frac{(4.14 \times 10^{-15} \text{ eV s})(3 \times 10^{18} \text{ angstrom/s})}{\lambda} = 4 \text{ eV} \]

### Step 6: Solve for Wavelength (λ)
Rearranging gives:
\[ \lambda = \frac{(4.14 \times 10^{-15})(3 \times 10^{18})}{4} \]

Calculating this:
1. Calculate \( 4.14 \times 3 = 12.42 \)
2. Then:
\[ \lambda = \frac{12.42 \times 10^{3}}{4} \]
\[ \lambda = 3.105 \times 10^{3} \text{ angstroms} \]

### Step 7: Final Answer
Thus, the wavelength of the incident light required to emit a photoelectron of zero velocity from the surface of the metal is approximately:
\[ \lambda \approx 3105 \text{ angstroms} \]