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In an atom electrons revolve around the nucleus along a path of radius $0.72 Å$ making $9.4 \times 10^{18}$ revolutions per second. The equivalent currents is [given, $e=1.6 \times 10^{-19} \mathrm{C}$ ]
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Verified Answer
The correct answer is:
$1.5 \mathrm{~A}$
Given, radius of circular path,
$$
\begin{aligned}
r & =0.72 Å=0.72 \times 10^{-10} \mathrm{~m}=7.2 \times 10^{-11} \mathrm{~m} \\
\left(\frac{n}{t}\right) & =9.4 \times 10^{18} \mathrm{rev} / \mathrm{s} . \\
e & =1.6 \times 10^{-19} \mathrm{C}
\end{aligned}
$$
$\therefore$ Equivalent current,
$$
I=\frac{n e}{t}=\left(\frac{n}{t}\right) e
$$
$$
=9.4 \times 10^{18} \times 1.6 \times 10^{-19}=1.504=1.5 \mathrm{~A}
$$
$$
\begin{aligned}
r & =0.72 Å=0.72 \times 10^{-10} \mathrm{~m}=7.2 \times 10^{-11} \mathrm{~m} \\
\left(\frac{n}{t}\right) & =9.4 \times 10^{18} \mathrm{rev} / \mathrm{s} . \\
e & =1.6 \times 10^{-19} \mathrm{C}
\end{aligned}
$$
$\therefore$ Equivalent current,
$$
I=\frac{n e}{t}=\left(\frac{n}{t}\right) e
$$
$$
=9.4 \times 10^{18} \times 1.6 \times 10^{-19}=1.504=1.5 \mathrm{~A}
$$
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