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Question
The radius of the shortest orbit in a one-electron system is 18 pm. It may be
Options
hydrogen
deuterium
He+
Li++
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Solution
Li++
The radius of the nth orbit in one electron system is given by
`r_n = (n^2a_0)/Z`
Here, a0 = 53 pm
For the shortest orbit,
n = 1
For hydrogen,
Z = 1
∴ Radius of the first state of hydrogen atom = 53 pm
For deuterium,
Z= 1
∴ Radius of the first state of deuterium atom = 53 pm
For He+,
Z = 2
∴ Radius of He+ atom =`53/2 pm = 26.5 "pm"`
For Li++,
Z = 3
∴ Radius of Li++ atom = `53/3 "pm" = 17.66"pm" ≈ 18 "pm"`
The given one-electron system having radius of the shortest orbit to be 18 pm may be Li++.
The given one-electron system having radius of the shortest orbit to be 18 pm may be Li++.
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