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, we will use the photoelectric effect equations derived from Einstein's photoelectric equation. The maximum kinetic energy of photoelectrons can be expressed as:

\[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{hc}{\lambda} - W \]

Where:
- \( K_{\text{max}} \) is the maximum kinetic energy of the photoelectrons.
- \( m \) is the mass of the electron.
- \( v_{\text{max}} \) is the maximum speed of the photoelectrons.
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)).
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).
- \( \lambda \) is the wavelength of the incident light.
- \( W \) is the work function of the metal.

### Step 1: Write the equations for both wavelengths

For the first wavelength \( \lambda_1 = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} \):

\[ \frac{1}{2} m v_1^2 = \frac{hc}{\lambda_1} - W \tag{1} \]

For the second wavelength \( \lambda_2 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \):

\[ \frac{1}{2} m v_2^2 = \frac{hc}{\lambda_2} - W \tag{2} \]

### Step 2: Divide equation (1) by equation (2)

\[ \frac{v_1^2}{v_2^2} = \frac{\frac{hc}{\lambda_1} - W}{\frac{hc}{\lambda_2} - W} \]

Given that the ratio of maximum speeds \( \frac{v_1}{v_2} = 3 \) (i.e., \( v_1 = 3v_2 \)), we have:

\[ \frac{v_1^2}{v_2^2} = \left(\frac{3}{1}\right)^2 = 9 \]

Thus, we can write:

\[ 9 = \frac{\frac{hc}{\lambda_1} - W}{\frac{hc}{\lambda_2} - W} \]

### Step 3: Cross-multiply and simplify

\[ 9 \left( \frac{hc}{\lambda_2} - W \right) = \frac{hc}{\lambda_1} - W \]

This simplifies to:

\[ 9 \frac{hc}{\lambda_2} - 9W = \frac{hc}{\lambda_1} - W \]

Rearranging gives:

\[ 9 \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = 8W \]

### Step 4: Substitute values for \( h \), \( c \), \( \lambda_1 \), and \( \lambda_2 \)

Substituting \( h = 6.626 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), \( \lambda_1 = 3000 \times 10^{-10} \, \text{m} \), and \( \lambda_2 = 6000 \times 10^{-10} \, \text{m} \):

\[ 9 \left( \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}} \right) - \left( \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} \right) = 8W \]

Calculating the left-hand side:

\[ 9 \left( \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}} \right) = 9 \left( \frac{6.626 \times 3}{6000} \times 10^{-34 + 8 + 10} \right) \]
\[ = 9 \left( \frac{19.878}{6000} \times 10^{-16} \right) = 2.9817 \times 10^{-16} \]

And for the second term:

\[ \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} = \frac{19.878}{3000} \times 10^{-16} = 6.626 \times 10^{-17} \]

Now substituting these into the equation:

\[ 2.9817 \times 10^{-16} - 6.626 \times 10^{-17} = 8W \]
\[ 2.319 \times 10^{-16} = 8W \]
\[ W = \frac{2.319 \times 10^{-16}}{8} = 2.89875 \times 10^{-17} \, \text{J} \]

Converting to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):

\[ W = \frac{2.89875 \times 10^{-17}}{1.6 \times 10^{-19}} \approx 1.81 \, \text{eV} \]

### Step 5: Calculate maximum speeds \( v_1 \) and \( v_2 \)

Now substituting \( W \) back into equation (1) to find \( v_1 \):

\[ \frac{1}{2} m v_1^2 = \frac{hc}{\lambda_1} - W \]

Substituting values:

\[ \frac{1}{2} (9.1 \times 10^{-31}) v_1^2 = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} - 1.81 \times 1.6 \times 10^{-19} \]

Calculating the right-hand side:

\[ \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} \approx 6.62 \times 10^{-19} \]
\[ 1.81 \times 1.6 \times 10^{-19} \approx 2.896 \times 10^{-19} \]
\[ \frac{1}{2} (9.1 \times 10^{-31}) v_1^2 = 6.62 \times 10^{-19} - 2.896 \times 10^{-19} = 3.724 \times 10^{-19} \]

Now solving for \( v_1 \):

\[ v_1^2 = \frac{2 \times 3.724 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 8.2 \times 10^{11} \]
\[ v_1 \approx 9 \times 10^5 \, \text{m/s} \]

For \( v_2 \):

\[ v_2 = \frac{v_1}{3} \approx 3 \times 10^5 \, \text{m/s} \]

### Final Answers
- Work function \( W \approx 1.81 \, \text{eV} \)
- Maximum speed \( v_1 \approx 9 \times 10^5 \, \text{m/s} \)
- Maximum speed \( v_2 \approx 3 \times 10^5 \, \text{m/s} \)