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Asked by Subhash Last Modified
In the Bohr model of the hydrogen atom, the radius of the electron orbit corresponding to the nn-th energy level (rnrn) is given by the formula:
rn=n2⋅h24π2⋅m⋅k⋅e2rn=4π2⋅m⋅k⋅e2n2⋅h2
where:
Given that the radius of the innermost electron orbit (corresponding to the ground state, n=1n=1) is 5.3×10−115.3×10−11 m, we can use this information to find the radius of the orbit in the second excited state (n=3n=3).
We can rewrite the formula for rnrn as follows:
rn=n2r1⋅r1rn=r1n2⋅r1
where:
Substituting the given values, we have:
r3=3212⋅(5.3×10−11 m)r3=1232⋅(5.3×10−11m)
r3=9×(5.3×10−11 m)r3=9×(5.3×10−11m)
r3=47.7×10−11 mr3=47.7×10−11m
r3=4.77×10−10 mr3=4.77×10−10m
Therefore, the radius of the orbit in the second excited state of the hydrogen atom is 4.77×10−104.77×10−10 m.
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