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The de Broglie wavelength of an electron...

The de Broglie wavelength of an electron with kinetic energy 120 e V is `("Given h"=6.63xx10^(-34)Js,m_(e)=9xx10^(-31)kg,1e V=1.6xx10^(-19)J)`

A

`2.13Ã…`

B

`1.13Ã…`

C

`4.15Ã…`

D

`3.14Ã…`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) : As `p=sqrt(2mK)`
`=sqrt(2xx9xx10^(-31)xx120xx1.6xx10^(-19))`
`=5.88xx10^(-24)kg" m s"^(-1)`
de Broglie wavelength,
`lamda=(h)/(p)=(6.63xx10^(-34))/(5.88xx10^(-24))=1.13xx10^(-10)m=1.13Ã…`
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