The de Broglie's wavelength of electron ...

The de Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :

A

`(0.529)/(2pi)Å`

B

`2pixx0.529Å`

C

`0.529Å`

D

`4xx0.529Å`

Verified by Experts

The correct Answer is:

B

`r_(n)=0.529(n)^(@)Å`

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