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The de-Broglie wavelength of an electron...

The de-Broglie wavelength of an electron revolving in the ground state orbit is

A

`pi r`

B

`pi r^(2)`

C

`2 pi r`

D

`sqrt(2 pi r)`

Text Solution

Verified by Experts

The correct Answer is:
C
`lambda=h/(mv)`
`lambda=(hr)/(mvr)=(hr)/((nh)/(2pi))`
`:. nlambda =2 pi r`
`:. lambda =2 pi r` (for ground state n = 1)
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