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If hydrogen atom, an electron jumps from bigger orbit to smaller orbit so that radius of smaller orbit is one-fourth of radius of bigger orbit. If speed of electron in bigger orbit was $\mathrm{v}$, then speed in smaller orbit is
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The correct answer is:
$2 v$
Radius of the orbit, $\mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^{2}$
$$
\begin{aligned}
& \frac{\mathrm{r}_{\mathrm{nbig}}}{\mathrm{r}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {big }}^{2}}{\mathrm{n}_{\text {small }}^{2}}=\frac{4}{1} \\
\Rightarrow & \frac{\mathrm{n}_{\text {big }}}{\mathrm{n}_{\text {small }}}=2 \\
\Rightarrow & \frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2}
\end{aligned}
$$
Velocity of electron in $\mathrm{n}^{\text {th }}$ orbit
$$
\begin{array}{c}
\quad \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\
\frac{\mathrm{v}_{\mathrm{n} \mathrm{big}}}{\mathrm{v}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2} \\
\Rightarrow \mathrm{v}_{\mathrm{n} \text { small }}=2\left(\mathrm{v}_{\mathrm{n} \mathrm{big}}\right)=2 \mathrm{v}
\end{array}
$$
$$
\begin{aligned}
& \frac{\mathrm{r}_{\mathrm{nbig}}}{\mathrm{r}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {big }}^{2}}{\mathrm{n}_{\text {small }}^{2}}=\frac{4}{1} \\
\Rightarrow & \frac{\mathrm{n}_{\text {big }}}{\mathrm{n}_{\text {small }}}=2 \\
\Rightarrow & \frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2}
\end{aligned}
$$
Velocity of electron in $\mathrm{n}^{\text {th }}$ orbit
$$
\begin{array}{c}
\quad \mathrm{v}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}} \\
\frac{\mathrm{v}_{\mathrm{n} \mathrm{big}}}{\mathrm{v}_{\mathrm{n} \text { small }}}=\frac{\mathrm{n}_{\text {small }}}{\mathrm{n}_{\text {big }}}=\frac{1}{2} \\
\Rightarrow \mathrm{v}_{\mathrm{n} \text { small }}=2\left(\mathrm{v}_{\mathrm{n} \mathrm{big}}\right)=2 \mathrm{v}
\end{array}
$$
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