Wave Nature Of Matter And De Broglie's Equation - Practice Questions & MCQ

As we know light behaves both as a wave and particle. Because If you are observing phenomenon like the interference, diffraction or reflection, you will find that light is a wave. However, if you are looking at phenomena like the photoelectric effect, you will find that light has a particle character.

De Broglie’s hypothesis stated that there is symmetry in nature and that if the light behaves as both particles and waves, matter too will have both the particle and wave nature.

i.e if lightwave can behave as a particle then the particle can also behave as waves.

De Broglie’s Equation-

According to De Broglie, A moving material particle can be associated with the wave.

De Broglie proposed that the wavelength  associated with the moving material particle of momentum p is given as 

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

where $h=$ plank's constant $^2$ and $h=6.626 \times 10^{-34} \mathrm{~J} s$
further, we can write De - Broglie wavelength as

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

where
$h=$ plank's constant
$m=$ mass of particle
$v=$ speed of the particle
$K=$ Kinetic energy of particle
So from De Broglie's Equation, we can conclude that-

$
\lambda \alpha \frac{1}{m}
$

i.e Wavelength associated with a heavier particle is smaller than that with a lighter particle.

$
\lambda \alpha \frac{1}{\nu}
$

i.e when Particle moves faster, then wavelength will be smaller and vice versa
- if the particle at rest then De - Broglie wavelength will be infinite $(\lambda=\infty)$

$
\lambda \alpha \frac{1}{p} \alpha \frac{1}{v} \alpha \frac{1}{\sqrt{K}}
$

- De - Broglie wavelength $(\lambda)$ is independent of charge.

De - Broglie wavelength of Electron-

As De Broglie's Equation is given as

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

So for an electron having velocity v attained by it when it is accelerated through a potential difference of V . then (Kinetic energy gain by the electron)=(work is done on an electron by the electric field)

$
\text { i.e } K=W_E \Rightarrow \frac{1}{2} m_e v^2=e V
$

So De - Broglie wavelength of Electron is given as

$
\lambda_e=\frac{h}{m_e v}=\frac{h}{\sqrt{2 m_e K}}=\frac{h}{\sqrt{2 m_e(e V)}}
$

using $h=6.626 \times 10^{-34} \mathrm{~J}$ and $m_e=9.1 \times 10^{-31} \mathrm{~kg}$ and $e=1.6 \times 10^{-19} \mathrm{C}$
we get $\lambda_e=\frac{12.27}{\sqrt{V}} A^{\circ} \quad$ (i.e answer will be in $A^0=$ Angstrom $^{\text {) }}$ )
Similarly, we can find De - Broglie wavelength associated with charged particle
De - Broglie wavelength with charged particle-

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

Where $K \rightarrow$ kinetic energy of particle
$q \rightarrow$ charged particle
$V \rightarrow$ potential diffenence
- De - Broglie wavelength of the proton

$
\text { using } m_p=1.67 \times 10^{-27} \mathrm{~kg}_{\text {and }} q_p=e=1.6 \times 10^{-19} \mathrm{C}
$
 

we get

$
\lambda_{\text {proton }}=\frac{0.286}{\sqrt{V}} A^{\circ}
$

- De - Broglie wavelength of Deuteron

$
\text { using } m_D=2 \times 1.67 \times 10^{-27} \mathrm{~kg}_{\text {and }} q_D=e=1.6 \times 10^{-19} \mathrm{C}
$

we get

$
\lambda_{\text {deutron }}=\frac{0.202}{\sqrt{V}} A^{\circ}
$

- De - Broglie wavelength of an Alpha particle $\left(\mathrm{He}^{2+}\right)$

$
\text { using } m_{\alpha^{2+}}=4 \times 1.67 \times 10^{-27} \mathrm{~kg}_{\text {and }} q_{\alpha^{2+}}=2 e=2 \times 1.6 \times 10^{-19} \mathrm{C}
$

we get

$
\lambda_{\alpha-\text { partical }}=\frac{0.101}{\sqrt{V}} A^{\circ}
$

- Electron microscope-

An electron microscope is an important application of de-Broglie waves designed to study very minute objects like viruses, microbes and the crystal structure of the solids. In the electron microscope, by selecting a suitable value of potential difference V , we can have an electron beam

$
\lambda_e=\frac{12.27}{\sqrt{V}} A^{\circ}
$