Dual Nature of matter and Radiation Part 2

Concept: The photoelectric effect relates the energy of incident light (frequency) to the work function and the maximum kinetic energy of the emitted photoelectrons.Key Rule/Formula: The energy of a photon is $E = h\nu = \frac{hc}{\lambda}$. The photoelectric equation is $K_{\text{max}} = h\nu - \phi$, or $K_{\text{max}} = \frac{hc}{\lambda} - \phi$.Solution/Analysis:1. According to the photoelectric effect, the maximum kinetic energy ($K$) of the emitted photoelectrons is given by $K = h\nu - \phi$, where $\phi$ is the work function and $h$ is Planck's constant.2. For light of wavelength $\lambda_1$, the energy is $E_1 = \frac{hc}{\lambda_1}$. Thus, $K_1 = \frac{hc}{\lambda_1} - \phi$.3. For light of wavelength $\lambda_2$, the energy is $E_2 = \frac{hc}{\lambda_2}$. Thus, $K_2 = \frac{hc}{\lambda_2} - \phi$.4. We are given the relationship between the wavelengths: $\lambda_1 = 2\lambda_2$. This implies $\; \frac{1}{\lambda_1} = \; \frac{1}{2\lambda_2}$.5. Substitute this relationship into the expression for $K_1$:$K_1 = \frac{hc}{\lambda_1} - \phi$$K_1 = \frac{hc}{2\lambda_2} - \phi$6. Now, we use the expression for $K_2$:$K_2 = \frac{hc}{\lambda_2} - \phi$We can rewrite $\frac{hc}{\lambda_2}$ as $2\left(\frac{hc}{2\lambda_2}\right)$. Substituting this into the equation for $K_2$:$K_2 = 2\left(\frac{hc}{2\lambda_2}\right) - \phi$7. Substitute the expression from step 5 into the equation for $K_2$:$K_2 = 2(K_1 + \phi) - \phi$ (This substitution is complex; let's use a simpler approach based on the structure of the provided solution.)Alternative Step-by-Step using the provided logic:1. From step 2 and 3, we have:$K_1 = \frac{hc}{\lambda_1} - \phi \;\; (i)$$K_2 = \frac{hc}{\lambda_2} - \phi \;\; (ii)$2. Using the given condition $\lambda_1 = 2\lambda_2$, we substitute $\lambda_1$ in equation $(i)$:$K_1 = \frac{hc}{2\lambda_2} - \phi \;\; (iii)$3. Rearrange equation $(iii)$ to find $\frac{hc}{2\lambda_2}$:$K_1 + \phi = \frac{hc}{2\lambda_2}$4. Now, substitute this expression into equation $(ii)$:$K_2 = \left(\frac{hc}{\lambda_2}\right) - \phi$Since $\frac{hc}{\lambda_2} = 2\left(\frac{hc}{2\lambda_2}\right)$, we get:$K_2 = 2(K_1 + \phi) - \phi$$K_2 = 2K_1 + 2\phi - \phi$$K_2 = 2K_1 + \phi$5. Rearranging the terms to find $\phi$:$\phi = K_2 - 2K_1$