Question

The velocity of an $e^{-}$ in excited state of H-atom is $1.093\times {10}^{6}m/s$. What is the circumference of this orbit?

• A. $3.32\times {10}^{-10}m$
• B. $6.64\times {10}^{-10}m$
• C. $13.30\times {10}^{-10}m$
• D. $13.28\times {10}^{-8}m$

Answer 1

Answer: (C)

$13.30\times {10}^{-10}m$

$v_n=2.186\times {10}^6\frac{Z}{n}$

$\Rightarrow 1.093\times {10}^6=2.186\times {10}^6\times \frac{1}{n}$; $n=2$

And we know,

$r=0.529\frac{n^2}{Z}\Rightarrow 0.529\times 4\mathring{A}$

$\therefore$ Circumference of the orbit

$=2\times \pi \times r=2\times \frac{22}{7}\times 0.529\times 4\times {10}^{-10}$

$\Rightarrow 13.30\times {10}^{-10}m$

Hence, the correct option is $C$