If speed of electron in ground state energy level is 2.2× 106 ms–1, then its speed in fourth excited state will be

A
6.8 × 10⁶ ms^–1
B
8.8 × 10⁵ ms^–1
C
5.5 × 10⁵ ms^–1
D
5.5 × 10⁶ ms^–1

Solution:

We know that $V = \frac{c}{137} \frac{Z}{n}; V \propto \frac{1}{n}$
$\begin{array}{l} \frac{V_{1}}{V_{2}} = \frac{n_{2}}{n_{1}}; \frac{2 .2 \times 10^{6}}{V_{2}} = \frac{4}{1} \\ V_{2} = \frac{2 .2 \times 10^{6}}{4} = 0 .55 \times 10^{6} = 5 .5 \times 10^{5} \end{array}$

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If speed of electron in ground state energy level is 2.2× 10⁶ ms^–1, then its speed in fourth excited state will be

Solution:

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