Given that a photon of light of wavelength `10,000A` has an energy equal to 1.23 eV. When light of wavelength `5000A` and intenstiy `I_0` falls on a photoelectric cell, the saturation current is `0.40xx10^(-6)A` and the stopping potential is 1.36V, then the work function is

Options

A. 0.43eV
B. 1.10 eV
C. 1.36 eV
D. 2.47 eV

Explanation

To solve the problem step by step, we will follow the principles of the photoelectric effect and the relationship between energy, wavelength, and work function.

### Step 1: Calculate the Energy of the Photon at Wavelength 5000 Å
We know that the energy of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) = Planck's constant \( = 4.14 \times 10^{-15} \, \text{eV s} \)
- \( c \) = Speed of light \( = 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) = Wavelength in meters

First, convert the wavelength from angstroms to meters:
\[ \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

Now, substituting the values into the energy formula:
\[ E_2 = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{5 \times 10^{-7} \, \text{m}} \]

Calculating:
\[ E_2 = \frac{(1.242 \times 10^{-6} \, \text{eV m})}{5 \times 10^{-7} \, \text{m}} = 2.484 \, \text{eV} \]

### Step 2: Use the Given Energy of the Photon at Wavelength 10000 Å
From the problem, we know that:
- The energy of a photon with wavelength 10,000 Å is given as 1.23 eV.
- Since the energy is inversely proportional to the wavelength, the energy of the photon at 5000 Å is double that of 10,000 Å.

Thus, we can confirm:
\[ E_2 = 2 \times 1.23 \, \text{eV} = 2.46 \, \text{eV} \]

### Step 3: Calculate the Kinetic Energy of the Emitted Electrons
The kinetic energy (KE) of the emitted electrons can be calculated using the stopping potential (V):
\[ KE = eV \]
Where:
- \( e \) = charge of electron \( = 1 \, \text{eV/V} \)
- \( V \) = stopping potential = 1.36 V

Thus,
\[ KE = 1.36 \, \text{eV} \]

### Step 4: Use Einstein's Photoelectric Equation to Find the Work Function
Einstein's photoelectric equation states:
\[ \phi = E - KE \]
Where:
- \( \phi \) = work function
- \( E \) = energy of the incident photon
- \( KE \) = kinetic energy of the emitted electrons

Substituting the known values:
\[ \phi = 2.46 \, \text{eV} - 1.36 \, \text{eV} \]
\[ \phi = 1.10 \, \text{eV} \]

### Conclusion
The work function \( \phi \) is \( 1.10 \, \text{eV} \).