Question

An electron is revolving in the n^th orbit of radius 4.2 Å, then the value of n is $(r_{1} = 0.529\text{ }\text{Å})$

• A. 4
• B. 5
• C. 6
• D. 3

Answer 1

Answer: (D)

3

Answer 2

Since $r_{n} \propto n^{2}$

$\frac{r_{n}}{r_{1}} = \frac{n^{2}}{(1)^{2}}\therefore r_{n} = n^{2}r_{1}$

Or $n^{2} = \frac{r_{n}}{r_{1}}$

$\Rightarrow n = \sqrt{\frac{r_{n}}{r_{1}}} = \sqrt{\frac{4.2}{0.529}} = \sqrt{7.939} = 2.81 = n^{2}$