JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 12)
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 :
$$\left(1.8 \times 10^{6}\right)$$
$$\left(7.5 \times 10^{6}\right)$$
$$\left(8.4 \times 10^{6}\right)$$
$$\left(3.6 \times 10^{6}\right)$$
Explanation
$$v\,\alpha {z \over n}$$
$${{{v_1}} \over {{v_2}}} = \left( {{{{n_2}} \over {{n_1}}}} \right)$$
$$ \Rightarrow {{3.6 \times {{10}^6}} \over {{v_2}}} = {3 \over 7}$$
$$ \Rightarrow {v_2} = {7 \over 3} \times 3.6 \times {10^6}$$ m/s
$$ = 8.4 \times {10^6}$$ m/s
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