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What is the angular momentum of an electron of de-Broglie wavelength ? Given : r is the radius of orbit -

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Step-by-step explanation:

The angular momentum (L) of an electron in an orbit can be calculated using the formula:

\[L = n \cdot \frac{h}{2\pi}\]

Where:

- \(n\) is the principal quantum number of the electron, which is related to the orbit it's in.

- \(h\) is the Planck constant, approximately \(6.62607015 × 10^{-34}\) J·s.

- \(2\pi\) is a constant.

The de Broglie wavelength (\(\lambda\)) of an electron can be related to the radius of the orbit (\(r\)) using the following equation:

\[\lambda = \frac{h}{p}\]

Where:

- \(p\) is the momentum of the electron.

Now, the momentum of the electron in an orbit can be related to the angular momentum and the radius of the orbit:

\[p = \frac{L}{r}\]

Substitute this expression for \(p\) back into the de Broglie wavelength equation:

\[\lambda = \frac{h}{L/r}\]

Simplify:

\[\lambda = \frac{hr}{L}\]

Now, you can solve for \(L\):

\[L = \frac{hr}{\lambda}\]

So, the angular momentum of an electron with a de Broglie wavelength \(\lambda\) in an orbit of radius \(r\) is \(L = \frac{hr}{\lambda}\).

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