Photoelectric Effects and deBroglie Equation: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Photoelectric Effects and deBroglie Equation. Understanding the underlying formula is key to arriving at the correct answer below:

Energy of incident light (assuming it was meant to be $500 nm$ for standard photoelectric emission since $5000 nm$ is only $0.248 eV$) $E = \frac{1240}{500} = 2.48 eV$. Maximum kinetic energy of emitted electron $K_{max} = E - \phi = 2.48 - 2.28 = 0.2 eV$. The minimum de-Broglie wavelength $\lambda_{min} = \frac{12.27}{\sqrt{0.2}} \AA \approx 27.4 \AA = 2.74 \times 10^{-9} m$. Therefore, the wavelength is $\geq 2.8 \times 10^{-9} m$.