Answer:
Information shadow:
n = 1, first orbit
Z = 2, for \[H{{e}^{+}}\]
Problem solving strategy:
\[E=-\frac{{{Z}^{2}}}{{{n}^{2}}}\times
2.18\times {{10}^{-18}}\text{J}\]
(Energy of hydrogen-like species)
\[r=\frac{{{n}^{2}}}{Z}\times
0.529\overset{{}^\circ }{\mathop{\text{A}}}\,\]
(Radius of nth orbit of hydrogen-like
species)
Working it out:
\[E=-\frac{{{2}^{2}}}{{{1}^{2}}}\times
2.18\times {{10}^{-18}}=-8.72\times {{10}^{-18}}J\]
\[r=\frac{{{1}^{2}}}{2}\times 0.529\overset{{}^\circ
}{\mathop{\text{A}}}\,=0.2645\overset{{}^\circ }{\mathop{\text{A}}}\,\]
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