1.

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]

Answer»

`3XX10^(-29)nm`
`6xx10^(-29)nm`
`3xx10^(29)nm`
`6xx10^(29)nm`

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


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