(a) De-Broglie put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions. If radiation shows dual aspects, so should matter. 

De-Broglie proposed that the wave length \(\lambda\) associated with a particle of momentum 'P' is given as:

\(\lambda\)= \(\frac{h}{P}\) = \(\frac{h}{mv}\) ; 

where,

m = mass of the particle  

v = particle speed

For a photon: P = \(\frac{hv}{c}\)

Therefore, \(\frac{h}{p}= \frac{c}{v}=\lambda\)

Thus, De-Broglie equation equals the  wavelength of em radiation of which the photon is a quantum of energy and momentum.

(b) De-broglie wavelength is given by: \(\lambda\) = \(\frac{h}{p}\)

\(\lambda\) = \(\frac{h}{\sqrt{2mqv}}\)

\(\frac{\lambda_d}{\lambda_{\alpha}}\) = \(\frac{\sqrt{m_{\alpha}q_{\alpha}}}{\sqrt{m_dq_d}}\) = \(\frac{\sqrt{4\times2}}{\sqrt{2\times1}}\) = 2

De-Broglie reasoned out that nature was symmetrical and two basic physical entities –mass and radiation must be symmetrical.If radiation shows shows dual aspect than matter should do so. 

De-Broglie equation- 

\(\lambda\)=\(\frac{h}{p}\)

For photon – 

P=\(\frac{hv}{C}\) 

Therefore,\(\frac{h}{P}=\frac{C}{v}=\lambda\) 

As \(\lambda=\frac{h}{\sqrt2mk}\) 

So,alpha particle will be having shortest de-Broglie wavelength compared to deutrons.