Question

derive de broglie wavelength associated with an electron is lamda=h/√2meV and lamda = h/√2mE​

Answer 1

Answer:

The de Broglie wavelength associated with an electron$\lambda=\frac{h}{\sqrt{2 m e V}}$ and $\lambda=\frac{h}{\sqrt{2 m E}}$ is derived.

Explanation:

Consider a charged particle of mass m carrying charge q which is accelerated from the rest through a potential V. The electric field produced does work on that particle. The work done on it will be equivalent to the kinetic energy E of the particle.

That is,

E=qV       ...(1)

qV is the work done on the charged particle.

Also,

The kinetic energy (E) of the charged particle can be expressed as,

$E=\frac{1}{2} mv^{2}$

or

$E = \frac{1}{2}\times mv\times v$       ...(2)

Now, the momentum of the charged particle is given by

$p = mv$                     ...(3)

Substitute equation (3) in equation (2) we get the expression for kinetic energy in terms of momentum,

$E=\frac{p^{2}}{2 m}$                    ...(4)

From the above equation, we can obtain another expression for momentum as,

$p = \sqrt{2Em}$               ...(5)  

Again substitute equation (1) in (5)  

$p = \sqrt{2qVm}$              ...(6)

For a charged particle, de Broglie wavelength λ will be,

λ$=\frac{h}{p}$                        ...(7)

After substituting the value of p using equation (5),

$\lambda=\frac{h}{\sqrt{2 m E}}$

Substitute equation (6) in (7)

$\lambda=\frac{h}{\sqrt{2 m q V}}$              ...(8)

For electron, charge $q = e$

So the equation(8) becomes,

$\lambda=\frac{h}{\sqrt{2 m e V}}$                  ...(9)

This is the expression for the de-Broglie wavelength associated with an electron.