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Step-by-Step Solution
Step 1: Identify the relevant physical principle
An electron moving in a circular orbit around the nucleus behaves like a moving charge producing a magnetic field at the center of its circular path. The magnetic field B at the center due to a charge q moving with speed v in a circular orbit of radius r is given by:
B = \frac{\mu_0}{4\pi} \cdot \frac{q \, v}{r^2} \,,
where \mu_0 / 4\pi = 10^{-7} \,\mathrm{N\,A^{-2}} and q is the charge of the electron.
Step 2: List the known quantities
Speed of the electron, v = 6.76 \times 10^6\,\mathrm{m\,s}^{-1}
Charge of the electron, e = 1.6 \times 10^{-19}\,\mathrm{C}
Radius of the orbit, r = 0.52\,\mathrm{\AA} = 0.52 \times 10^{-10}\,\mathrm{m}
\displaystyle \frac{\mu_0}{4\pi} = 10^{-7}\,\mathrm{N\,A^{-2}}
Step 3: Substitute values into the formula
B = 10^{-7} \times \frac{\bigl(1.6 \times 10^{-19}\bigr) \times \bigl(6.76 \times 10^6\bigr)}{\bigl(0.52 \times 10^{-10}\bigr)^2}.
Step 4: Calculate the denominator r^2
r^2 = \bigl(0.52 \times 10^{-10}\bigr)^2
= 0.52^2 \times 10^{-20}
= 0.2704 \times 10^{-20}
= 2.704 \times 10^{-21}\,\mathrm{m}^2.
Step 5: Evaluate the entire expression
1. Numerator:
10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^6
= (1.6 \times 6.76) \times 10^{-19+6-7}
= 10.816 \times 10^{-20}
= 1.0816 \times 10^{-19}.
2. Denominator:
2.704 \times 10^{-21}.
Therefore,
B = \frac{1.0816 \times 10^{-19}}{2.704 \times 10^{-21}}
\approx 40\,\mathrm{T}.
Step 6: State the final answer
The magnetic field produced at the nucleus by the revolving electron in the hydrogen atom is approximately 40 T.
