Q179 / 200 · Manipal (MET) 2011 · Physics

The velocity of an electron in the second orbit of sodium atom (atomic number $=11$ ) is $v$. The velocity of an electron in its fifth orbit will be

A
$v$
B
$\frac{22}{5} v$
C
$\frac{5}{2}$
Correct answer
D
$\frac{2}{5} v$

Solution

Velocity of electron in $n$th orbit $\begin{aligned} v_n & =\frac{Z e^2}{2 \varepsilon_0 n h} \\ \Rightarrow \quad v & \propto \frac{1}{n} \\ v_1 & =v, n_1=2, n_2=5\end{aligned}$ $\therefore \quad \frac{v}{v_2}=\frac{n_2}{n_1}$ $\therefore \quad \frac{v}{v_2}=\frac{5}{2}$ $\Rightarrow \quad v_2=\frac{2 v}{5}$

Hence, the correct answer is option: 'D'.

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