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(D) The number of spectral lines $N$ obtained due to transitions of electrons from the $n$th orbit to lower orbits is given by the formula: $N = \frac

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$3/4$

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(D) The number of spectral lines $N$ obtained due to transitions of electrons from the $n$th orbit to lower orbits is given by the formula: $N = \frac{n(n-1)}{2}$.For the first case,$N = 6$:$6 = \frac{n_1(n_1-1)}{2} \Rightarrow n_1^2 - n_1 - 12 = 0 \Rightarrow (n_1-4)(n_1+3) = 0$. Since $n > 0$,we have $n_1 = 4$.For the second case,$N = 3$:$3 = \frac{n_2(n_2-1)}{2} \Rightarrow n_2^2 - n_2 - 6 = 0 \Rightarrow (n_2-3)(n_2+2) = 0$. Since $n > 0$,we have $n_2 = 3$.The velocity of an electron in the $n$th orbit of a hydrogen atom is given by $v_n \propto \frac{1}{n}$.Therefore,the ratio of the velocities is:$\frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{3}{4}$.

An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocities of the electron in these two orbits is

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