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JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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The radius of Bohr's first orbit in H-atom is 0.053 nm. The radius of second orbit in He+\mathrm{H}{\mathrm{e}^ + } would be:
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Correct option: C
0.1060 nm
Option C (0.1060 nm) is correct because the radius of the nth Bohr orbit is proportional to n²/Z, where n is the principal quantum number and Z is the atomic number.

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