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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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Answer» Correct Answer - 4 `n=(lamda)/(2pir_(1))=1340/(2xx3.14xx53)=4` |
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