MHT CET · Physics · Laws of Motion
If an electron is revolving around the hydrogen nucleus at a distance of \(0.1 \mathrm{~nm}\), what should be its speed?
- A \(2.188 \times 10^{6} \mathrm{~ms}^{-1}\)
- B \(1.094 \times 10^{6} \mathrm{~ms}^{-1}\)
- C \(4.376 \times 10^{6} \mathrm{~ms}^{-1}\)
- D \(1.59 \times 10^{6} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(1.59 \times 10^{6} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Electrostatic force \(=\) centripetal force
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r^{2}}=\frac{m v^{2}}{r}\)
\(\therefore \quad v=\sqrt{\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{m r}\right)}\)
\(=\sqrt{\frac{9 \times 10^{9} \times 1 \times\left(1.6 \times 10^{-19}\right)^{2}}{\left(9.1 \times 10^{-31}\right) \times\left(0.1 \times 10^{-9}\right)}}=1.59 \times 10^{6} \mathrm{~ms}\)
\(-1\)
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r^{2}}=\frac{m v^{2}}{r}\)
\(\therefore \quad v=\sqrt{\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{m r}\right)}\)
\(=\sqrt{\frac{9 \times 10^{9} \times 1 \times\left(1.6 \times 10^{-19}\right)^{2}}{\left(9.1 \times 10^{-31}\right) \times\left(0.1 \times 10^{-9}\right)}}=1.59 \times 10^{6} \mathrm{~ms}\)
\(-1\)
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