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If linear momentum of a particle is 2.2x...

If linear momentum of a particle is `2.2xx10^(4) kgms^(-1)` then what will be its de-Broglie wavelength ? `[h=6.6xx10^(-34)Js]`

A

`3xx10^(-29)nm`

B

`6xx10^(-29)nm`

C

`3xx10^(29)nm`

D

`6xx10^(29)nm`

Text Solution

Verified by Experts

The correct Answer is:
A

`lambda =(h)/(p)=(6.6xx10^(-34))/(2.2xx10^(4))=3xx10^(-38)m`
`lambda=(3xx10^(-38))/(10^(-9))=3xx10^(-29)nm`
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