A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are `mu_1` and mu_2` respectively. If the ratio `u_1: u_2 = 2:1 and hc= 1240 eV` nm, the work function of the metal is neraly. (a)3.7 eV (b) 3.2 eV (c ) 2.8eV (d) 2.5eV.

Explanation

To solve the problem, we will use the photoelectric effect equation, which relates the kinetic energy of the emitted photoelectrons to the energy of the incident photons and the work function of the metal.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy of Photoelectrons**:
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be expressed as:
\[ K.E. = \frac{hc}{\lambda} - \phi \]
where \( \phi \) is the work function of the metal, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Setting Up the Equations**:
For the first wavelength \( \lambda_1 = 248 \, \text{nm} \):
\[ K.E._{1} = \frac{hc}{\lambda_1} - \phi \]
For the second wavelength \( \lambda_2 = 310 \, \text{nm} \):
\[ K.E._{2} = \frac{hc}{\lambda_2} - \phi \]

3. **Substituting Known Values**:
Given \( hc = 1240 \, \text{eV} \cdot \text{nm} \):
\[ K.E._{1} = \frac{1240}{248} - \phi = 5 \, \text{eV} - \phi \]
\[ K.E._{2} = \frac{1240}{310} - \phi = 4 \, \text{eV} - \phi \]

4. **Relating Kinetic Energies to Velocities**:
The kinetic energy can also be expressed in terms of the maximum speeds \( u_1 \) and \( u_2 \):
\[ K.E._{1} = \frac{1}{2} m u_1^2 \]
\[ K.E._{2} = \frac{1}{2} m u_2^2 \]
Given the ratio \( \frac{u_1}{u_2} = 2 \), we can express \( u_1 \) as \( 2u_2 \).

5. **Substituting the Velocity Ratio**:
From the velocity ratio:
\[ \frac{u_1^2}{u_2^2} = \left(\frac{u_1}{u_2}\right)^2 = 4 \]
Thus, we can write:
\[ \frac{K.E._{1}}{K.E._{2}} = \frac{5 - \phi}{4 - \phi} = 4 \]

6. **Setting Up the Equation**:
Setting the two expressions equal:
\[ 5 - \phi = 4(4 - \phi) \]
Expanding this gives:
\[ 5 - \phi = 16 - 4\phi \]

7. **Solving for the Work Function**:
Rearranging the equation:
\[ 4\phi - \phi = 16 - 5 \]
\[ 3\phi = 11 \]
\[ \phi = \frac{11}{3} \approx 3.67 \, \text{eV} \]

8. **Final Answer**:
Rounding to one decimal place, the work function \( \phi \) is approximately \( 3.7 \, \text{eV} \).

### Conclusion:
The work function of the metal is nearly \( 3.7 \, \text{eV} \).