Question

Calculate the ratio of $4^{\mathrm{th}}$ Bohr's to $2^{\mathrm{nd}}$ Bohr's orbit of ${\mathrm{He}}^{+}$ion.

Answer 1

Step 1: Given data and concept used

$4^{\mathrm{th}}$ Bohr’s to $2^{\mathrm{nd}}$ Bohr’s orbit of ${\mathrm{He}}^{+}$ ion

n^th Bohr’s orbit radius is

${\mathrm{r}}_{\mathrm{n}}=0.529\frac{{\mathrm{n}}^2}{\mathrm{z}}{\mathrm{A}}^0$

Step 2:Calculating the ratio of ${\mathrm{He}}^{+}$ion

For ${\mathrm{He}}^{+}$,$\mathrm{z}=2,\mathrm{n}=2$

According to the formula of radius $\mathrm{r}=0.529\frac{{\mathrm{n}}^2}{\mathrm{z}}$

The atomic number of ${\mathrm{He}}^{+}$ ion is 2.

Thus, $\mathrm{z}=2$

Since, it is $4^{\mathrm{th}}$ orbit , $\mathrm{n}=2$

$4^{\mathrm{th}}$ Bohr’s orbit of ${\mathrm{He}}^{+}$ ion is

$0.529\frac{{\left(4\right)}^2}{2}$.$=\frac{0.529\times 16}{2}$

$2^{\mathrm{nd}}$ Bohr’s orbit of ${\mathrm{He}}^{+}$ ion is

$0.529\frac{{\left(2\right)}^2}{2}$$=\frac{0.529\times 4}{2}$

The ratio of ${\mathrm{He}}^{+}$ ion is $=\frac{\frac{0.529\times 16}{2}}{\frac{0.529\times 4}{2}}$

$=\frac{0.529\times 16}{2}\times \frac{2}{0.529\times 4}$

$=\frac{4}{1}$

$=4:1$

Therefore, The ratio of $4^{\mathrm{th}}$ Bohr’s to $2^{\mathrm{nd}}$ Bohr’s orbit of ${\mathrm{He}}^{+}$ ion is 4:1