Question

For any particle, its particle nature and wave nature can be observed simultaneously. Comment. Radiation of wavelength \( \lambda \) is incident on a surface of negligible work function. Find the de-Broglie wavelength of the emitted electron.

Hint:

Wave and particle natures are complementary, never seen in one experiment. With \( \phi \approx 0 \), all photon energy \( hc/\lambda \) becomes kinetic energy, so \( \lambda' = h/\sqrt{2m\,KE} = \sqrt{h\lambda/(2mc)} \).

Answer 1

Step 1: Why both natures are not seen together.
Light and material particles show a dual character, but the two faces never appear in one and the same measurement. A which-path (particle) measurement destroys the interference (wave) pattern, and observing interference means we cannot tell which path the quantum took. This mutual exclusivity is Bohr's complementarity principle, so the given claim is false.

Step 2: All photon energy goes to the electron.
Since the surface has negligible work function, no energy is spent in freeing the electron. The photon energy \(hc/\lambda\) is converted entirely into the electron's kinetic energy: \(\tfrac{1}{2}mv^2 = \dfrac{hc}{\lambda}\).

Step 3: Speed of the emitted electron.
Solving for \(v\): \(v = \sqrt{\dfrac{2hc}{m\lambda}}\).

Step 4: Apply the de-Broglie relation.
\(\lambda' = \dfrac{h}{mv} = \dfrac{h}{m\sqrt{2hc/(m\lambda)}} = \dfrac{h}{\sqrt{2mhc/\lambda}}\).

Step 5: Reduce to the simplest form.
Squaring inside the root, \(\lambda' = \sqrt{\dfrac{h^2\lambda}{2mhc}} = \sqrt{\dfrac{h\lambda}{2mc}}\).

\[\boxed{\lambda' = \sqrt{\dfrac{h\lambda}{2mc}}}\]