Atomic StructureHard

Question

The speed of electron revolving around H-nucleus is 0.547 × 106 m/s. The distance of electron from the nucleus is

Options

A.2.116 Å
B.4.761 Å
C.8.464 Å
D.0.529 Å

Solution

$V_{n} = 2.188 \times 10^{6}\frac{z}{n}\text{m/s}$

$\Rightarrow 0.547 \times 10^{6} = 2.188 \times 10^{6} \times \frac{1}{n} $$$\therefore n = 4$$

Now, $r_{n} = 0.529 \times \frac{n^{2}}{z} = 0.529 \times \frac{4^{2}}{1} = 8.464\overset{o}{A}$

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