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The de-Broglie wavelength of a particle ...

The de-Broglie wavelength of a particle having a momentum of `2xx10^(-28)kg-ms^(-1)` is

A

`3.3xx10^(-5) m`

B

`6.6xx10^(-6) m`

C

`3.3xx10^(-6) m`

D

`1.65xx10^(-6)m`

Text Solution

Verified by Experts

The correct Answer is:
C
`lambda=h/p=(6.63xx10^(-34))/(2xx10^(-28))=3.3xx10^(-6) m`.
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