Derive an expression for de Brogile wave length?

Answer 1

The derivation of De Brogile's wave length:

We know that the energy of light E=mc^2

Again if the frequency of light is f

Then E=h*f [where h is planck constant]

So,

mc^2=h*f

or, mc^2=h*(c/l) [where c is the velocity of light,l is the wavelength ,& c=l*f]

or,l=(h*c)/mc^2

or,l=h/m*c

or,l=h/p.[ where p is the momentum=m*c]

this is the De Broglie's equation.

That derivation is completely wrong. The momentum of a photon is NOT given by m*c, as c=constant, m=0. Neither is the energy of light mc^2, as that also goes to zero. The derivation works by a miracle.

The proper derivation is this:

From a relativistic invariant we know that

E^2-p^2c^2=(mc^2)^2

For a photon, the mass = 0, so

E/c=p

For a photon,

E=hc/lambda

Therefore,

h/lambda=p

The first step works as a result of contra/covariant vectors, but can be checked using algebra. The relativistic Energy = m*c^2/sqrt(1-v^2/c^2) , momentum p = m*v/sqrt(1-v^2/c^2).

Answer 2

Louis de Broglie proposed that particles, such as electrons, exhibit both wave-like and particle-like properties. The wavelength of such a particle is given by the de Broglie wavelength and can be calculated using the following formula:

�=ℎ�

λ=

p

h

​

where:

�

λ is the de Broglie wavelength,

ℎ

h is Planck's constant (

6.626×10−34

6.626×10

−34

J·s),

�

p is the momentum of the particle.

For a particle with mass

�

m and velocity

�

v, the momentum (

�

p) is given by the product of mass and velocity:

�=�⋅�

p=m⋅v

So, you can substitute this expression for

�

p into the de Broglie wavelength formula:

�=ℎ��

λ=

mv

h

​

This formula relates the de Broglie wavelength (

�

λ) of a particle to its mass (

�

m) and velocity (

�

v). Keep in mind that this concept is particularly important in the realm of quantum mechanics, where particles can exhibit both wave and particle properties.

Answer 3

The de Broglie wavelength (λ) is given by the equation λ = h/p, where h is the Planck constant (6.626 x 10^-34 J·s) and p is the momentum of the particle. In terms of mass (m) and velocity (v), the momentum can be expressed as p = mv.

Answer 4

http://www.dpedtech.com/VelocityEquation.pdf

see page 3

Answer 5

De Broglie's wavelength is expressed as h/mv where h= Planck's constant (6.63x10-34) and mv=momentum (or mass x velocity )