Question

Derive an expression for the De Broglie wavelength of an electron under a potential difference V

Answer 1

Answer: this is the derivation for de Broglie relationship chemistry

Explanation:

Answer 2

DERIVATION IS GIVEN BELOW

We know , according to De broglie's hypothesis :

$p=\dfrac{h}{\lambda}$  .

Now, Kinetic energy of an electron in potential difference V is = eV.

e -   charge on electron.

Also, $K.E=\dfrac{p^2}{2m}.$  ( p is momentum)

Therefore, $eV=\dfrac{p^2}{2m}$

Putting value of p from  De broglie's equation :

We get ,

$eV=\dfrac{\dfrac{h^2}{\lambda^2}}{2m}$

Arranging them all :

We get,

$\lambda=\dfrac{h}{\sqrt{2 \times m \times eV}}$

Hence , this is the required solution.

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Modern Physics

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