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The wavelength lambda of de Broglie wave...

The wavelength `lambda` of de Broglie waves associated with an electron (mass `m`, charge) accelerated through a potential difference of `V` is given by (`h` is planck's constant):

A

`lambda=h//mV`

B

`lambda=h//2meV`

C

`lambda=h//sqrt(meV)`

D

`lambda=h//sqrt(2meV)`

Text Solution

Verified by Experts

The correct Answer is:
D

`lambda=h/P=h/sqrt(2mK)=h/sqrt(2meV)`
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