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The electron in the first excited state ...

The electron in the first excited state `(n_(1)=2)` of `H`- atom absorbs a photon and is further excited to `n^(th)` shell. The de-Broglie wavelength of the electron in this excited state is 1340 pm. Find the value of `n`.

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The correct Answer is:
4
`n=(lamda)/(2pir_(1))=1340/(2xx3.14xx53)=4`
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