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de-Broglie wavelength of an electron in ...

de-Broglie wavelength of an electron in the nth Bohr orbit is `lambda_(n)` and the angular momentum is `J_(n)` then

A

(a)`J_(n) prop lambda_(n)`

B

(b)`lambda_(n) prop (1)/(J_(n))`

C

(c)`lambda_(n) prop J_(n)^(2)`

D

(d)None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`J_(n)=(nh)/(2pi) implies n lambda_(n)=2 pi (r_(0) n^(2))`
So `J_(n) prop lambda_(n) implies lambda_(n) =(2pi r_(0))n`
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