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Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is:

1. 6.6 Å
2. 3.5 Å
3. 9.7 Å
4. 12.9 Å

Correct answer: 9.7 Å

Solution

Second excited state means n=3, r=4.65 A. de-Broglie wavelength lambda = 2*pi*r/n = 2*pi*4.65/3 = 9.7 A. So the answer is 9.7 A.

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