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The de-Broglie wavelength of an electron...

The de-Broglie wavelength of an electron in the first Bohr orbit is

A

`2px`

B

`6px`

C

`9x`

D

`x//3`

Text Solution

Verified by Experts

The correct Answer is:
B
`r_(n) = n^(2) r_(1),`
`mv_(n) r_(n) = (nh)/(2pi)`
`lambda= (h)/(mv_n)`
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