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(B) The speed of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $v_n = \frac{v_1}{n}$,where $v_1$ is the speed in the ground state $

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$5.5 	imes 10^5 \ m/s$

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(B) The speed of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $v_n = \frac{v_1}{n}$,where $v_1$ is the speed in the ground state $(n=1)$.Given,$v_1 = 2.2 	imes 10^6 \ m/s$.The third excited state corresponds to $n = 4$ (since ground state is $n=1$,first excited is $n=2$,second is $n=3$,and third is $n=4$).Using the relation $v_n = \frac{v_1}{n}$,we get:$v_4 = \frac{v_1}{4} = \frac{2.2 	imes 10^6 \ m/s}{4}$.$v_4 = 0.55 	imes 10^6 \ m/s = 5.5 	imes 10^5 \ m/s$.

If the speed of an electron of a hydrogen atom in the ground state is $2.2 \times 10^6 \ m/s$,then its speed in the third excited state will be

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