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The de Broglie wavelength of an electron...

The de Broglie wavelength of an electron is 66 nm. The velocity of the electron is `(h=6.6xx10^(-34)kgm^(2)s^(-1)m=9.0xx10^(-31)kg)`

A

`1.84xx10^(-14)ms^(-1)`

B

`1.1xx10^(-4)ms^(-1)`

C

`5.4 xx 10^(3)ms^(-1)`

D

`1.1xx10^(4)ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`lamda=h/(mv)=(6.6xx10^(-34))/(9.0xx10^(-31)xx66xx10^(-9))`
`=1.1xx10^(4)ms^(-1)`
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