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What is the significance of de Broglie’s hypothesis in modern physics?

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Asked by Pallavi Arora

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Answered by

Chandra Pruthi

4 months ago

Significance of de Broglie's Hypothesis in Modern Physics (NCERT-based answer):

De Broglie's hypothesis is a revolutionary idea which changed the understanding of matter and wave as we know it. This hypothesis introduced the concept that particles of matter, like electrons, and Proton, can behave like wave and have wave-like properties, just as light exhibits both wave and particle nature.

According to de Broglie, any moving particle has a wavelength given by:

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

where $\lambda$ is the wavelength, $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity of the particle.

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A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wave-train of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wave train (in cm) when it reaches the top of the rope?

alok kumar singh

At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm

At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?

Since frequency (f) remains the same:
f = v? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm

In the Young's double slit experiment, the distance between the slits varies in time as d(t) = d? + a? sin(ωt); where d?, ω and a? are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

Raj Pandey

β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)

An unpolarised light beam of intensity 2l₀ is passed through a polaroid P and then through another polaroid Q which is oriented is such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is

Payal Gupta

$I = I_{0} c o s^{2} 30 °$

$= I_{0} {(\frac{\sqrt[]{3}}{2})}^{2} = \frac{3}{4} I_{0}$

In Young’s double slit experiment the two slits are 0.6mm distance apart. Interference pattern is observed on a screen at a distance 80cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be__________nm.

Vishal Baghel

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\Rightarrow \frac{d y}{D} = (2 n + 1) \frac{λ}{2}$ [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{d y}{D} = \frac{λ}{2}$

$λ = \frac{2 d}{D} y$

$= \frac{2 d}{D} \times \frac{d}{2} [y = \frac{d}{2}, G i v e n$ first dark fringe is observed on the screen directly opposite to one of the slits]

$λ = \frac{2 \times 0.6 \times 0.6}{800 \times 2}$

$λ = 450 m m$

In a Young's double slit experiment, the separation between the slits is $0.15 m m$ . In the experiment, a source of light of wavelength $589 n m$ is used and the interference pattern is observed on a screen kept $1.5 m$ away. The separation between the successive bright fringes on the screen is:

alok kumar singh

$\frac{Energy}{Volume} = \frac{{M L}^{2} T^{- 2}}{L^{3}} = ({M L}^{- 1} T^{- 2})$

The distance between two successive bright fringes is fringe width $(β)$ .

$β = \frac{λ D}{d} = \frac{589 \times 10^{- 9} \times 1.5}{0.15 \times 10^{- 3}} = 5.9 m m$

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