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 wavelength of the second source is

Explanation

To solve the problem step-by-step, we will use the photoelectric effect equation and the information provided in the question.

### Step 1: Understand the Photoelectric Effect Equation
The photoelectric effect can be described by the equation:
\[ E = hf = \frac{hc}{\lambda} - W \]
Where:
- \( E \) is the energy of the emitted photoelectrons,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( f \) is the frequency of the light,
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light,
- \( W \) is the work function of the material.

### Step 2: Set Up the Equations for Each Light Source
For the first light source with a wavelength of \( 4950 \, \text{Å} \):
- The critical potential is \( V_1 = 0.6 \, \text{V} \).
- The energy of the emitted electrons can be expressed as:
\[ eV_1 = \frac{hc}{\lambda_1} - W \]
Substituting \( \lambda_1 = 4950 \, \text{Å} = 4950 \times 10^{-10} \, \text{m} \):
\[ 0.6e = \frac{hc}{4950 \times 10^{-10}} - W \]
(Equation 1)

For the second light source:
- The critical potential is \( V_2 = 1.1 \, \text{V} \).
- The energy of the emitted electrons can be expressed as:
\[ eV_2 = \frac{hc}{\lambda_2} - W \]
(Equation 2)

### Step 3: Subtract the Two Equations
Subtract Equation 1 from Equation 2:
\[ eV_2 - eV_1 = \left(\frac{hc}{\lambda_2} - W\right) - \left(\frac{hc}{\lambda_1} - W\right) \]
This simplifies to:
\[ e(V_2 - V_1) = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \]
Substituting \( V_2 = 1.1 \, \text{V} \) and \( V_1 = 0.6 \, \text{V} \):
\[ e(1.1 - 0.6) = \frac{hc}{\lambda_2} - \frac{hc}{4950 \times 10^{-10}} \]
\[ 0.5e = \frac{hc}{\lambda_2} - \frac{hc}{4950 \times 10^{-10}} \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ \frac{hc}{\lambda_2} = 0.5e + \frac{hc}{4950 \times 10^{-10}} \]

### Step 5: Solve for \( \frac{1}{\lambda_2} \)
Now, we can express this as:
\[ \frac{1}{\lambda_2} = \frac{0.5e}{hc} + \frac{1}{4950 \times 10^{-10}} \]

### Step 6: Substitute Values
Substituting the values:
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Calculating:
\[ \frac{0.5 \times 1.6 \times 10^{-19}}{6.626 \times 10^{-34} \times 3 \times 10^8} + \frac{1}{4950 \times 10^{-10}} \]

### Step 7: Calculate \( \lambda_2 \)
After performing the calculations, we find:
\[ \lambda_2 \approx 4125 \, \text{Å} \]

### Final Answer
The wavelength of the second source is approximately \( 4125 \, \text{Å} \).

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