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

`J_(n) prop lambda_(n)`

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

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

None of these

Verified by Experts

The correct Answer is:

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

Equal to one fourth the circumference of the first orbit

Equal to half the circyumference of the first orbit

Equal to twice the circumference of the first orbit

Equal to the circulference of the first orbit

`2pia_0`

`4pia_0`

`6pia_0`

`8pia_0`

`J_n alpha lamda_n`

`lamda_n alpha 1//J_n`

`J_n alpha sqrtlamda_n`

None of these