De Broglie Relationship - Practice Questions & MCQ

De-broglie wavelength

(1) de-Broglie proposed that just like light, matter should exhibit both particle and wave-like properties. This means that just as the photon has momentum as well as wavelength, electrons should also have momentum as well as wavelength, and he proposed the following mathematical relationship:

$\lambda=\frac{h}{m v}=\frac{h}{p}=\frac{h}{\sqrt{2 m K E}}=\frac{h}{\sqrt{2 m q V}}$

where m is the mass of the particle, v its velocity, p its momentum,

KE is the Kinetic Energy of the particle,

V is the voltage across which the Charged particle having charge q is accelerated.

(2) de Broglie’s prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction, a phenomenon characteristic of waves.

(3) It needs to be noted that according to de Broglie, every object in motion has a wave character. This wavelength is quite significant for the subatomic particles which have very small masses. The wavelengths associated with ordinary objects are however so short  that their wave properties cannot be detected as they have large masses. 

(4) Bohr's model and the de Broglie's relation : Number of standing waves made by an electron in n^th Bohr orbit

According to Bohr's model, 

$m v r=\frac{n h}{2 \pi}$

According to de Broglie's Relation ,

$p=\frac{h}{\lambda}$

Combining the two

$2 \pi r=n \lambda$

So, the number of waves made by any electron in the nth orbit is equal to the principal quantum number of the orbit i.e. n.