Speed of an electron in Bohr's $7^{\text {th }}$ orbit for Hydrogen atom is $3.6 \times 10^{6} \mathrm{~m} / \mathrm{s}$. The corresponding speed of the electron in $3^{\text {rd }}$ orbit, in $\mathrm{m} / \mathrm{s}$ is :
Solution
<p>$v\,\alpha {z \over n}$</p>
<p>${{{v_1}} \over {{v_2}}} = \left( {{{{n_2}} \over {{n_1}}}} \right)$</p>
<p>$\Rightarrow {{3.6 \times {{10}^6}} \over {{v_2}}} = {3 \over 7}$</p>
<p>$\Rightarrow {v_2} = {7 \over 3} \times 3.6 \times {10^6}$ m/s</p>
<p>$= 8.4 \times {10^6}$ m/s</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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