The de Broglie wavelength associated with the kinetic energy K is \(\lambda=\frac{h}{\sqrt{2 m k}}\)
Where m is the mass of the particle
Since proton and electron have same KE, the wavelength is inversely proportional to square root of the mass \(\lambda \alpha \frac{1}{\sqrt{m}}\)
Mass of proton is 1840 times greater them that of electron. Therefore, de Broglie's wavelength of electron is greater than the proton.

According to de Broglie hypothesis, if radiation has a dual nature, then the moving particles of matter Iike electrons, protons, neutrons in motion should exhibit wave like character under an appropriate conditions. These waves are called de Broglie waves or matter waves.

For a given metallic Surface, the emission of photo electrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.

\(\text { (i) } \lambda_{\mathrm{d}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV_1}}} ; \lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{\mathrm{me}}}\)
\(\frac{\lambda_{\mathrm{d}}}{\lambda_{\alpha}}=\frac{\frac{1}{\sqrt{2 m e}}}{\frac{1}{\sqrt{4 m 2 e}}} \Rightarrow \lambda_{\mathrm{d}}=2 \lambda_{\alpha}\)
\(\text { (ii) } =\frac{K.E_d}{K.E_a}=\frac{eV}{2eV} =\frac{1}{2}\Rightarrow K.E_d=\frac{1}{2}K.E_a\)
∴ K.E of deuteron is half of K.E of α-particle.

(i) Davisson - Germer experiment
(ii) Diffraction phenomenon