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For the Bohr's first orbit of circumfere...

For the Bohr's first orbit of circumference ` 2pi r` , the de - Broglie wavelength of revolving electron will be

A

` 2 pi r `

B

`pir `

C

`(1)/( 2pir )`

D

`(1)/(4 pir)`

Text Solution

Verified by Experts

The correct Answer is:
A
`mvr=(nh)/(2pi)`, according to Bohr's theory
`implies 2pir=n((h)/(mv))=n lambda ` for ` n=1, lambda=2pir `
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