A photon incident on a metal of photo electric work function 2 ev produced photo electron of maximum kinetic energy 2 ev. The wave length associated with the photon is

Options

A. `6200A^(@)`
B. `3100A^(@)`
C. `9300A^(@)`
D. `2000A^(@)`

Answer

To solve the problem step by step, we will use the principles of the photoelectric effect and the relationship between energy, wavelength, and frequency of photons.

### Step 1: Understand the Photoelectric Equation
The maximum kinetic energy (K.E.) of the emitted photoelectron can be expressed using Einstein's photoelectric equation:
\[ K.E. = E_P - \phi \]
where:
- \( K.E. \) is the maximum kinetic energy of the emitted electron,
- \( E_P \) is the energy of the incident photon,
- \( \phi \) is the work function of the metal.

### Step 2: Substitute Given Values
From the problem, we know:
- Work function \( \phi = 2 \, \text{eV} \)
- Maximum kinetic energy \( K.E. = 2 \, \text{eV} \)

Substituting these values into the equation:
\[ K.E. = E_P - \phi \]
\[ 2 \, \text{eV} = E_P - 2 \, \text{eV} \]

### Step 3: Solve for Photon Energy
Rearranging the equation gives:
\[ E_P = K.E. + \phi \]
\[ E_P = 2 \, \text{eV} + 2 \, \text{eV} = 4 \, \text{eV} \]

### Step 4: Relate Energy to Wavelength
The energy of a photon can also be expressed in terms of its wavelength (\( \lambda \)):
\[ E_P = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

### Step 5: Rearranging for Wavelength
From the equation above, we can rearrange to find the wavelength:
\[ \lambda = \frac{hc}{E_P} \]

### Step 6: Substitute Values
We will use the value of \( E_P \) we found:
- Convert \( E_P = 4 \, \text{eV} \) to Joules (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E_P = 4 \, \text{eV} = 4 \times 1.6 \times 10^{-19} \, \text{J} = 6.4 \times 10^{-19} \, \text{J} \]

Now substituting the values of \( h \) and \( c \):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{6.4 \times 10^{-19} \, \text{J}} \]

### Step 7: Calculate Wavelength
Calculating the above expression:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{6.4 \times 10^{-19}} \]
\[ \lambda \approx 3.1 \times 10^{-7} \, \text{m} \]

### Step 8: Convert to Angstroms
To convert meters to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 3.1 \times 10^{-7} \, \text{m} = 3100 \, \text{angstroms} \]

### Final Answer
The wavelength associated with the photon is approximately:
\[ \lambda \approx 3100 \, \text{angstroms} \]

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