NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #10The radius of the shortest orbit in a one-electron system is 18 pm. It may be
(a) hydrogen
(b) deuterium
(c) He+
(d) Li++digAnsr: dAns : (d) Li++
The radius of the nth orbit in one electron system is given by
`` {r}_{\,\mathrm{\,n\,}}=\frac{{n}^{2}{a}_{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 = `` \frac{53}{2}\,\mathrm{\,pm\,}=26.5\,\mathrm{\,pm\,}``
For Li++,
Z = 3
∴ Radius of Li++ atom = `` \frac{53}{3}\,\mathrm{\,pm\,}=17.66\,\mathrm{\,pm\,}\approx 18\,\mathrm{\,pm\,}``
The given one-electron system having radius of the shortest orbit to be 18 pm may be Li++.
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