Question

De-broglie wavelength of electron.

Answer 1

Hey there!

Answer:


Consider that an electron of mass 'm' and charge 'e' is accelerated through a potential difference 'V'. If 'E' is the energy acquired by the particle, then:

E = eV

If 'v' is the velocity of the electron, then:

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

v = $\sqrt{\dfrac{2E}{m}}$

Let the above equation the first equation.

Now, de - broglie wavelength of electron is given by:

λ = $\dfrac{h}{mv}$

Using the first equation:

λ = $\dfrac{h}{m\sqrt{2E/m}}$

Or,

λ = $\dfrac{h}{\sqrt{2mE}}$

Substituting the value of E = eV

λ = $\dfrac{h}{m\sqrt{2meV}}$


That is the required equation of de - broglie wavelength of electron.



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Answer 2

${\huge{\textbf{\underline{De-broglie wavelength of electron}}}}$

de Broglie reasoned that matter also can show wave-particle duality, just like light, since light can behave both as a wave (it can be diffracted and it has a wavelength) and as a particle (it contains packets of energy hν). And also reasoned that matter would follow the same equation for wavelength as light namely,

➞λ = h / p

Where p is the linear momentum, as shown by Einstein.

$\bf Derivation :-$

de Broglie derived the above relationship as follows:

1) E = hν for a photon and λν = c for an electromagnetic wave.

2) E = mc2, means λ = h/mc, which is equivalent to λ = h/p.

Note: m is the relativistic mass, and not the rest mass; since the rest mass of a photon is zero.

Now, if a particle is moving with a velocity v, the momentum p = mv and hence λ = h / mv

Therefore, the de Broglie wavelength formula is expressed as;

➞λ = h / mv