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If the Planck's constant is h=6.6xx10^(-...

If the Planck's constant is `h=6.6xx10^(-34)J` , the de Broglie wavelength of a particle having momentum of `3.3xx10^(-24)kgms^(-1)` will be

A

`0.02Å`

B

`0.5Å`

C

`2Å`

D

`500Å`

Text Solution

Verified by Experts

The correct Answer is:
C
`lamda=h/(mv)=(6.6xx10^(-34)Js)?(3.3xx10^(-24)kgms^(-1))`
`=2xx10^(-10)m=2Å`
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