In the second excited state of a hydrogen atom, the velocity of electron in the orbit will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → ~0.73 × 106 m/s
In the Bohr model the speed of the electron in hydrogen is inversely proportional to n. The second excited state is n = 3 (ground state n = 1, first excited state n = 2).
\( \displaystyle v_n = \frac{e^{2}}{2\varepsilon_0 h}\times\frac{1}{n} = \frac{v_1}{n} \)
\( \displaystyle v_1 = \frac{(1.6\times10^{-19})^{2}}{2 \times 8.85\times10^{-12} \times 6.63\times10^{-34}} \approx 2.19\times10^{6}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle v_3 = \frac{2.19\times10^{6}}{3} \)
\( \displaystyle v_3 \approx 0.73\times10^{6}\,\mathrm{m\,s^{-1}} \)