Calculate Work Function Using Kinetic Energy of Photoelectrons

According to Einstein's photoelectric equation, the maximum kinetic energy $K$ of photoelectrons emitted from a metal surface is given by: $K = \frac{hc}{\lambda} - \phi$

where:

• $h$ is Planck's constant,
• $c$ is the speed of light in vacuum,
• $\lambda$ is the wavelength of the incident light,
• $\phi$ is the work function of the material.

For the two given light wavelengths $\lambda_1$ and $\lambda_2$, the respective maximum kinetic energies $K_1$ and $K_2$ are:

$K_1 = \frac{hc}{\lambda_1} - \phi \quad \text{--- (1)}$

$K_2 = \frac{hc}{\lambda_2} - \phi \quad \text{--- (2)}$

We are given the relation between the wavelengths: $\lambda_1 = 2\lambda_2 \implies \lambda_2 = \frac{\lambda_1}{2}$

Substituting $\lambda_2 = \frac{\lambda_1}{2}$ into equation (2): $K_2 = \frac{hc}{\left(\frac{\lambda_1}{2}\right)} - \phi$ $K_2 = \frac{2hc}{\lambda_1} - \phi \quad \text{--- (3)}$

From equation (1), we can express $\frac{hc}{\lambda_1}$ as: $\frac{hc}{\lambda_1} = K_1 + \phi$

Now, substitute this value into equation (3): $K_2 = 2(K_1 + \phi) - \phi$ $K_2 = 2K_1 + 2\phi - \phi$ $K_2 = 2K_1 + \phi$

Solving for the work function $\phi$: $\phi = K_2 - 2K_1$

Thus, the work function of the material is given by $K_2 - 2K_1$.

Correct Option: D ($K_2 - 2K_1$)