Question

Ratio of radius of first orbit of Li2+to the third orbit of He+ion will be

Answer 1

Answer: Here is the answer

Explanation:

Answer 2

Final answer: Ratio of radius of first orbit of $Li^{2+}$to the third orbit of $He^{+}$ion will be $\frac{2}{27}$.

Given that: We are given $Li^{2+}$ and $He^{+}$.

To find: We have to find the ratio of radius of first orbit of $Li^{2+}$to the third orbit of $He^{+}$ion.

Explanation:

• Radius of $n^{th}$ Bohr orbit $= r_{n} = \frac{n^{2}h^{2}}{4\pi^{2}mZe^{2}} = \frac{0.0529n^{2}}{Z}$ $nm$

Where,

n = Number of orbit  

Z =  atomic number of atom

• Radius of first orbit of $Li^{2+}$:

For $Li^{2+}$, atomic number Z = 3

First orbit n = 1

Radius of first orbit of $Li^{2+}$ = $r_{1} = \frac{0.0529 * 1^{2}}{3} = \frac{0.0529}{3}$ $nm$

• Radius of third orbit of $He^{+}$:

For $He^{+}$, atomic number Z = 2

Third orbit n = 3

Radius of third orbit of $He^{+}$ = $r_{3} = \frac{0.0529 * 3^{2}}{2} = \frac{0.0529*9}{2}$ $nm$

• Ratio of radius of first orbit of $Li^{2+}$to the third orbit of $He^{+}$ion

                                                          $=\frac{r_{1} of Li^{2+}}{r_{3} of He^{+}}$

                                                          $={\frac{(\frac{0.0529}{3})}{\frac{(0.0529*9}{2})}$

                                                          $=\frac{0.0529}{3}*\frac{2}{0.0529*9}\\\\=\frac{2}{3*9}\\\\ =\frac{2}{27}$

• Hence, ratio of radius of first orbit of $Li^{2+}$to the third orbit of $He^{+}$ion = $\frac{2}{27}$

To know more about the concept please go through the links

https://brainly.in/question/12516745

https://brainly.in/question/9568573

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