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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

`pi r`

B

`2 pi r`

C

`1/(2pi r)`

D

`1/(4 pi r)`

Text Solution

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The correct Answer is:
B
`lambda=h/(mv)`
`lambda=(hr)/(mvr)=(hr)/((nh)/(2pi))`
`n lambda =(2pi r)/n" "(n=1)`
`:. lambda =2 pi r`
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