KCET · Physics · Current Electricity
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, \(e=1.6 \times 10^{-19} \mathrm{C}\) ]
- A \(1.5 \mathrm{~A}\)
- B \(1.4 \mathrm{~A}\)
- C \(1.8 \mathrm{~A}\)
- D \(1.2 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(1.5 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Given, radius of circular path,
\(r =0.72 Å=0.72 \times 10^{-10} \mathrm{~m}=7.2 \times 10^{-11} \mathrm{~m} \)
\(\left(\frac{n}{t}\right) =9.4 \times 10^{18} \mathrm{rev} / \mathrm{s} . \)
\(e =1.6 \times 10^{-19} \mathrm{C}\)
\(\therefore\) Equivalent current,
\(I=\frac{n e}{t}=\left(\frac{n}{t}\right) e\)
\(=9.4 \times 10^{18} \times 1.6 \times 10^{-19}=1.504=1.5 \mathrm{~A}\)
\(r =0.72 Å=0.72 \times 10^{-10} \mathrm{~m}=7.2 \times 10^{-11} \mathrm{~m} \)
\(\left(\frac{n}{t}\right) =9.4 \times 10^{18} \mathrm{rev} / \mathrm{s} . \)
\(e =1.6 \times 10^{-19} \mathrm{C}\)
\(\therefore\) 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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