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The de-Broglie wavelength of a particle ...

The de-Broglie wavelength of a particle of charge q and mass m is `lamda`. If its kinetic energy be K, then

A

`lamda=(h)/(sqrt(mK))`

B

`lamda=(h)/(sqrt(mqK))`

C

`lamda=(sqrt(2mK))/(h)`

D

`lamda=(sqrt(mqK))/(h)`.

Text Solution

Verified by Experts

The correct Answer is:
A
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