The radius of orbit of an electron and the sp | vedclass

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$(A)$ Given for ground state $(n=1)$: $r_1 = 5.5 	imes 10^{-11} \,m$ and $v_1 = 4 	imes 10^6 \,m/s$. For the first excited state,$n=2$. Since $r_n \

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$6.908 	imes 10^{-16} \,s$

Vedclass · Answer

$(A)$ Given for ground state $(n=1)$: $r_1 = 5.5 	imes 10^{-11} \,m$ and $v_1 = 4 	imes 10^6 \,m/s$. For the first excited state,$n=2$. Since $r_n \propto n^2$,the radius $r_2 = r_1 	imes 2^2 = 5.5 	imes 10^{-11} 	imes 4 = 22.0 	imes 10^{-11} \,m$. Since $v_n \propto \frac{1}{n}$,the speed $v_2 = \frac{v_1}{2} = \frac{4 	imes 10^6}{2} = 2 	imes 10^6 \,m/s$. The orbital period $T$ is given by $T = \frac{2 \pi r_2}{v_2}$. Substituting the values: $T = \frac{2 	imes 3.14159 	imes 22.0 	imes 10^{-11}}{2 	imes 10^6} = 6.911 	imes 10^{-16} \,s$. Rounding to the nearest option,we get $6.908 	imes 10^{-16} \,s$.

The radius of orbit of an electron and the speed of an electron in the ground state of a hydrogen atom are $5.5 \times 10^{-11} \,m$ and $4 \times 10^6 \,m/s$ respectively. Then,the orbital period of this electron in the first excited state will be

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