Question

If \[\left( {\dfrac{{0.51 \times {{10}^{ - 10}}}}{4}} \right)m\] is the radius of the smallest electron orbit in Hydrogen like atom, then this atom is
(A) \[H - atom\]
(B) \[H{e^ + }\]
(C) \[L{i^{2 + }}\]
(D) \[B{e^{3 + }}\]

Answer 1

Hint:The problem is solved by using Bohr’s model of an atom which says electrons in an atom are located at a definite distance from the nucleus and are revolving around it with definite velocity.

Formula used:Radius of Hydrogen like atom is given by \[{r_n} =
\dfrac{{{n^2}}}{Z}{r_0}\]
 \[n = \] number of orbits which take only positive integer values \[i.e{\text{ }}n = 1,2,3,......\]
\[{r_0} \approx 0.51 \times {10^{ - 10}}m\], is called Bohr radius which is the smallest allowed radius for the hydrogen atom\[Z\] is the atomic number of atoms.

Complete step by step answer:
For smallest electron orbit \[n{\text{ }} = 1,\] the by putting the values in above formula we get
\[\left( {\dfrac{{0.51 \times {{10}^{ - 10}}}}{4}} \right) = \dfrac{{{1^2}}}{Z} \times 0.51 \times {10^{ - 10}}m\]
=>\[Z = \dfrac{{0.51 \times {{10}^{ - 10}} \times 4}}{{0.51 \times {{10}^{ - 10}}}}\]
\[\therefore Z = 4\]
$Z = 4$, which is the atomic number of Beryllium. The no of electron present in Beryllium is \[4.\] When it gives away \[3\] electron it acquires \[ + 3\] positive charge with one electron like hydrogen \[i.e\] \[B{e^{3 + }}\]
Hence the answer is option \[\left( D \right)\] \[B{e^{3 + }}\]

Note:Bohr’s Model of the Hydrogen atom isn’t applicable for atoms having more than one electron. It's only applicable for Hydrogen like atoms. The value of Bohr radius \[({r_{0}}) \approx 0.529 \times {10^{ - 10}}m\]. We have used the \[{r_0} \approx 0.51 \times {10^{ - 10}}m\] for simplification of the problem, as the answer remains the same up to few decimal places. Bohr’s Model of the Hydrogen atom isn’t applicable for atoms having more than one electron. It's only applicable for Hydrogen like atoms. The value of Bohr radius \[({r_{0)}}) \approx 0.529 \times {10^{ - 10}}m\]. We have used the \[{r_0} \approx 0.51 \times {10^{ - 10}}m\] for simplification of the problem, as the answer remains the same up to few decimal places.