A metallic surface is illuminated alternatively with light of wavelenghts `3000 Å` and `6000 Å`. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1 . Calculate the work function of the metal and the maximum speed of the photoelectrons in two cases.

Explanation

To solve the problem step by step, we will use the photoelectric effect and Einstein's photoelectric equation.

### Step 1: Write Einstein's Photoelectric Equation
According to Einstein's photoelectric equation, the maximum kinetic energy (K.E.) of the photoelectrons can be expressed as:
\[ K.E. = \frac{hc}{\lambda} - \phi \]
where:
- \( K.E. \) is the kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the metal.

### Step 2: Write the Equations for Both Wavelengths
For the first wavelength (\( \lambda_1 = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} \)):
\[ \frac{1}{2} m V_1^2 = \frac{hc}{\lambda_1} - \phi \quad \text{(1)} \]

For the second wavelength (\( \lambda_2 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)):
\[ \frac{1}{2} m V_2^2 = \frac{hc}{\lambda_2} - \phi \quad \text{(2)} \]

### Step 3: Divide the Two Equations
Dividing equation (1) by equation (2):
\[ \frac{V_1^2}{V_2^2} = \frac{\frac{hc}{\lambda_1} - \phi}{\frac{hc}{\lambda_2} - \phi} \]
Given that the ratio of the maximum speeds is \( \frac{V_1}{V_2} = 3 \), we have:
\[ \frac{V_1^2}{V_2^2} = \frac{9}{1} \]
Thus, we can write:
\[ 9 = \frac{\frac{hc}{\lambda_1} - \phi}{\frac{hc}{\lambda_2} - \phi} \]

### Step 4: Cross Multiply and Rearrange
Cross-multiplying gives:
\[ 9 \left( \frac{hc}{\lambda_2} - \phi \right) = \left( \frac{hc}{\lambda_1} - \phi \right) \]
Expanding this:
\[ 9 \frac{hc}{\lambda_2} - 9\phi = \frac{hc}{\lambda_1} - \phi \]
Rearranging gives:
\[ 9 \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = 8\phi \]

### Step 5: Substitute the Values
Substituting \( h = 6.63 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ 9 \left( \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}} \right) - \left( \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} \right) = 8\phi \]

Calculating:
\[ \frac{hc}{\lambda_2} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}} = 3.315 \times 10^{-19} \, \text{J} \]
\[ \frac{hc}{\lambda_1} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} = 6.63 \times 10^{-19} \, \text{J} \]

Substituting these values:
\[ 9(3.315 \times 10^{-19}) - (6.63 \times 10^{-19}) = 8\phi \]
Calculating:
\[ 29.835 \times 10^{-19} - 6.63 \times 10^{-19} = 8\phi \]
\[ 23.205 \times 10^{-19} = 8\phi \]
\[ \phi = \frac{23.205 \times 10^{-19}}{8} = 2.900625 \times 10^{-19} \, \text{J} \approx 1.81 \, \text{eV} \]

### Step 6: Calculate Maximum Speeds
Now substituting \( \phi \) back into equation (1) to find \( V_1 \):
\[ \frac{1}{2} m V_1^2 = \frac{hc}{\lambda_1} - \phi \]
Using \( m = 9.1 \times 10^{-31} \, \text{kg} \):
\[ \frac{1}{2} (9.1 \times 10^{-31}) V_1^2 = 6.63 \times 10^{-19} - 2.900625 \times 10^{-19} \]
Calculating:
\[ \frac{1}{2} (9.1 \times 10^{-31}) V_1^2 = 3.729375 \times 10^{-19} \]
\[ V_1^2 = \frac{2 \times 3.729375 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 8.2 \times 10^{11} \]
\[ V_1 \approx 9.06 \times 10^5 \, \text{m/s} \]

For \( V_2 \):
\[ V_2 = \frac{V_1}{3} \approx 3.02 \times 10^5 \, \text{m/s} \]

### Final Answers
- Work function \( \phi \approx 1.81 \, \text{eV} \)
- Maximum speed \( V_1 \approx 9.06 \times 10^5 \, \text{m/s} \)
- Maximum speed \( V_2 \approx 3.02 \times 10^5 \, \text{m/s} \)