The work function for a metal is 4 eV. To eject the photo electrons with zero velocity the wavelength of the incident light should be

Options

A. `2700Å`
B. `1700Å`
C. `5900Å`
D. `3100^(@)A`

Answer

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

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of sufficient energy strikes a metal surface, it can eject electrons from that surface. The energy of the incident light must be equal to or greater than the work function of the metal to eject electrons.

### Step 2: Set Up the Energy Equation
The energy of the incident photons can be expressed as:
\[ E = h \nu \]
where \( E \) is the energy of the photon, \( h \) is Planck's constant, and \( \nu \) is the frequency of the light.

When the electrons are ejected with zero velocity, all the energy of the photon is used to overcome the work function (\( \phi \)) of the metal:
\[ h \nu = \phi \]

### Step 3: Substitute the Work Function
Given that the work function \( \phi \) is 4 eV, we can write:
\[ h \nu = 4 \text{ eV} \]

### Step 4: Relate Frequency to Wavelength
We know that the speed of light \( c \) is related to frequency and wavelength by the equation:
\[ c = \lambda \nu \]
where \( \lambda \) is the wavelength of the light. Rearranging this gives:
\[ \nu = \frac{c}{\lambda} \]

### Step 5: Substitute Frequency in the Energy Equation
Substituting \( \nu \) in the energy equation gives:
\[ h \frac{c}{\lambda} = 4 \text{ eV} \]

### Step 6: Solve for Wavelength
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = \frac{hc}{4 \text{ eV}} \]

### Step 7: Use Known Values
We know:
- Planck's constant \( h = 4.135667696 \times 10^{-15} \text{ eV s} \)
- Speed of light \( c = 3 \times 10^8 \text{ m/s} \)

To convert the units appropriately, we can use:
\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]
and \( 1 \text{ m} = 10^{10} \text{ Å} \).

### Step 8: Calculate the Wavelength
Substituting the values:
\[ \lambda = \frac{(4.135667696 \times 10^{-15} \text{ eV s})(3 \times 10^8 \text{ m/s})}{4 \text{ eV}} \]

Calculating this gives:
\[ \lambda = \frac{1.240 \times 10^{-6} \text{ eV m}}{4 \text{ eV}} = 3.1 \times 10^{-7} \text{ m} = 3100 \text{ Å} \]

### Final Answer
The wavelength of the incident light should be **3100 Å**.

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