The de Broglie wavelength () associated with a photoelectron varies with the frequency () of the incident radiation as…

🧠 The Surplus Energy Wave The de Broglie wavelength of an electron ($\lambda$) is determined by its momentum. That momentum comes from the "leftover" energy after the photon pays the metal's threshold price ($u_0$).

🗺️ The Square Root Derivation

1. Kinetic Energy ($KE$): Using Einstein's equation: $KE = h(\nu - \nu_0)$.
2. De Broglie Wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot KE}}$
3. Combining the Two: $\lambda = \frac{h}{\sqrt{2m \cdot h(\nu - \nu_0)}}$ $\lambda \propto \frac{1}{\sqrt{\nu - \nu_0}} = (\nu - \nu_0)^{-1/2}$

⚡ The "Energy-Wavelength" Bridge Remember the core link: $\lambda \propto 1/\sqrt{KE}$. Since Energy is linear with frequency ($u$), the wavelength must be inversely proportional to the square root of the frequency surplus.

⚠️ Common Traps Don't pick the direct inverse ($1/x$). The square root in the momentum formula ($p = \sqrt{2mE}$) is the defining feature of particle wavelengths.

$\boxed{\text{Answer: (d)}}$