An electron in the hydrogen atom circles around the proton with a speed of \(2.18 \times 10^6\) m/s in an orbit of radius \(0.53\,\text{Å}\). The equivalent current is:

Hydrogen Atom Electron Current – Numerical Solution

Question: An electron in the hydrogen atom circles around the proton with a speed of \(2.18 \times 10^6\) m/s in an orbit of radius \(0.53\,\text{Å}\). The equivalent current is:

a) \(1.048 \times 10^{-2}\,\text{A}\)

b) \(1.048 \times 10^{-3}\,\text{A}\)

c) \(1.048 \times 10^{-4}\,\text{A}\)

d) \(1.058 \times 10^{-4}\,\text{A}\)

Detailed Solution:

The equivalent current due to an electron circling with frequency \(f\) is

\[ I = f\,e, \quad\text{with}\quad f = \frac{v}{2\pi r}. \]

Given:

• \(v = 2.18 \times 10^6\ \mathrm{m/s}\)
• \(r = 0.53\ \mathrm{Å} = 0.53 \times 10^{-10}\ \mathrm{m}\)
• \(e = 1.6 \times 10^{-19}\ \mathrm{C}\)

Compute the frequency:

\[ f = \frac{2.18 \times 10^6}{2\pi \times 0.53 \times 10^{-10}} \approx 1.17 \times 10^{13}\ \mathrm{s^{-1}}. \]

Thus the current is:

\[ I = f\,e = \bigl(1.17 \times 10^{13}\bigr)\times\bigl(1.6\times10^{-19}\bigr) \approx 1.048\times10^{-3}\ \mathrm{A}. \]

Correct answer: (b) \(1.048 \times 10^{-3}\,\text{A}\)