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: Comment on the statement.
The statement is not correct. According to the principle of complementarity, the wave nature and the particle nature of matter (or radiation) are complementary aspects. In any single experiment only one of them can be observed: an experiment designed to show wave behaviour cannot at the same time show particle behaviour, and vice versa. So the two natures cannot be observed simultaneously.

Step 2: Energy of the incident photon.
A photon of the incident radiation of wavelength \( \lambda \) has energy \( E = \dfrac{hc}{\lambda} \).

Step 3: Kinetic energy of the emitted electron.
By Einstein's photoelectric equation \( KE = E - \phi \), where \( \phi \) is the work function. Here \( \phi \) is negligible (\( \phi \approx 0 \)), so the whole photon energy becomes kinetic energy:
\( KE = \dfrac{hc}{\lambda} \).

Step 4: de-Broglie wavelength of the electron.
The de-Broglie wavelength is \( \lambda' = \dfrac{h}{p} \), where the momentum \( p = \sqrt{2m\,KE} = \sqrt{\dfrac{2mhc}{\lambda}} \).
Therefore \( \lambda' = \dfrac{h}{\sqrt{2mhc/\lambda}} \).

Step 5: Simplify.
\[ \lambda' = \sqrt{\dfrac{h^2}{2mhc/\lambda}} = \sqrt{\dfrac{h^2\lambda}{2mhc}} = \sqrt{\dfrac{h\lambda}{2mc}} \]

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