Photoelectric Effects and deBroglie Equation: Practice Problem & Solution

Solution Explained:

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 radiation $E = 13.6 (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times \frac{3}{4} = 10.2 eV$. The stopping potential is $3.57 V$, so $K_{max} = 3.57 eV$. Work function $W = E - K_{max} = 10.2 - 3.57 = 6.63 eV$. Using $W = h\nu_0$, we get threshold frequency $\nu_0 = \frac{6.63 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} Hz$.