When a surface is irradiated with light of wavelength `4950 Å`, a photocurrent appears which vanishes if a retarding potential greater than 0.6 volt is applied across the phototube. When a second source of light is used, it is found that the critical potential is changed to 1.1 volt.
The work- function of the emitting surface is i

Options

A. `2.2eV`
B. `1.5eV`
C. `1.9eV`
D. `1.1eV`

Explanation

To find the work function of the emitting surface, we can use the photoelectric effect equation, which relates the stopping potential, the energy of the incident photons, and the work function of the material. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Wavelength of light, \( \lambda = 4950 \, \text{Å} = 4950 \times 10^{-10} \, \text{m} \)
- Stopping potential (retarding potential), \( V_0 = 0.6 \, \text{V} \)

### Step 2: Use the Photoelectric Equation
The photoelectric effect can be described by the equation:
\[ E_k = h\nu - \phi \]
where:
- \( E_k \) is the kinetic energy of the emitted electrons,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the material.

The kinetic energy of the emitted electrons can also be expressed in terms of the stopping potential:
\[ E_k = eV_0 \]
where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 3: Relate Frequency to Wavelength
The frequency \( \nu \) can be calculated using the speed of light \( c \):
\[ \nu = \frac{c}{\lambda} \]
Substituting \( c = 3 \times 10^8 \, \text{m/s} \) and \( \lambda = 4950 \times 10^{-10} \, \text{m} \):
\[ \nu = \frac{3 \times 10^8}{4950 \times 10^{-10}} \approx 6.06 \times 10^{14} \, \text{Hz} \]

### Step 4: Substitute Values into the Photoelectric Equation
Now, substituting \( E_k \) and \( \nu \) into the photoelectric equation:
\[ eV_0 = h\nu - \phi \]
Rearranging gives:
\[ \phi = h\nu - eV_0 \]

### Step 5: Calculate \( h\nu \)
Now calculate \( h\nu \):
\[ h\nu = h \cdot \frac{c}{\lambda} = 6.626 \times 10^{-34} \cdot \frac{3 \times 10^8}{4950 \times 10^{-10}} \approx 3.94 \times 10^{-19} \, \text{J} \]

### Step 6: Calculate \( eV_0 \)
Now calculate \( eV_0 \):
\[ eV_0 = 1.6 \times 10^{-19} \cdot 0.6 = 9.6 \times 10^{-20} \, \text{J} \]

### Step 7: Find the Work Function \( \phi \)
Now substitute \( h\nu \) and \( eV_0 \) into the equation for \( \phi \):
\[ \phi = 3.94 \times 10^{-19} - 9.6 \times 10^{-20} = 2.98 \times 10^{-19} \, \text{J} \]

### Step 8: Convert Work Function to Electron Volts
To convert the work function from joules to electron volts:
\[ \phi_{eV} = \frac{\phi}{e} = \frac{2.98 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.86 \, \text{eV} \]

### Final Answer
The work function of the emitting surface is approximately \( 1.86 \, \text{eV} \).

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