2nd excited state implies that n=3. Now the inner most radius i.e., for n=1, r'= 5.3*10^-11 m Now we know radius of nth orbit is given by r(n) = r' * n^2 There the radius of the second excited state i.e., r(n =3) = r' * 3^2 = 5.3*10^-11*9 = 4.77*10^-10 m. I hope my answer helps. Have a great day!
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Answer – We are given the radius of the innermost orbit of a hydrogen atom as r1 = 5.3 × 10−11 m. Suppose r2 represents the radius of the orbit for the level n = 2. Then, It will related to the radius of the inner most orbit as: \[{{r}_{2}}={{n}^{2}}{{r}_{1}}=4\times 5.3\times {{10}^{-11}}m\] \[{{r}_{2}}=2.12\times {{10}^{-10}}m\] For n = 3, the corresponding electron radius is given by – \[{{r}_{3}}=9\times 5.3\times {{10}^{-11}}=4.77\times {{10}^{-10}}m\] Therefore, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10−10 m and 4.77 × 10−10 m respectively. |
If the radius of the first orbit of hydrogen atom is 5.3 into 10 to the power minus 11 m
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