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Which state of the triply ionized `Be^(+++)` has the same orbital radius as that of the ground state of hydrogen? Compare the energies of two states. |
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Answer» Correct Answer - `2;4` Suppose orbital radius of groung state of hydrogen (n=1) is `r_(1)` Radius of nth orbit of hydrogen like atom is `r_(n)=(n^(2))/Z r_(1)` for `Be^(+++), Z=4` As `r_(n)=r_(1)`, therefore, `(n^(2))/4=1 or n=2` Let energy of electrons in ground state of hydrogen `=E_(1)`. Energy of electron in nth state of hydrogen like atom `E=(Z^(2))/(n^(2)) E_(1)=(4^(2))/(2^(2)) E_(1)=4E_(1)` `:. (E_(2))/(E_(1))=4` |
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