1.

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

Answer»

`0.02Å`
`0.5Å`
`2Å`
`500Å`

Solution :`lamda=h/(MV)=(6.6xx10^(-34)JS)?(3.3xx10^(-24)kgms^(-1))`
`=2xx10^(-10)m=2Å`


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