The difference between the radii of n^{text{t | vedclass

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(B) The radius of the $n^{	ext{th}}$ orbit in a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius. Thus,$r_n \propto n^2$.Acco

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$\frac{2h}{\pi}$

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(B) The radius of the $n^{	ext{th}}$ orbit in a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius. Thus,$r_n \propto n^2$.According to the problem,the difference between the radii of the $(n+1)^{	ext{th}}$ and $n^{	ext{th}}$ orbits is equal to the radius of the $(n-1)^{	ext{th}}$ orbit:$r_{n+1} - r_n = r_{n-1}$Substituting the proportionality $r_n \propto n^2$:$(n+1)^2 - n^2 = (n-1)^2$$n^2 + 2n + 1 - n^2 = n^2 - 2n + 1$$2n = n^2 - 2n$$n^2 - 4n = 0$Since $n 
eq 0$,we have $n = 4$.According to Bohr's quantization postulate,the angular momentum $L$ of an electron in the $n^{	ext{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.For $n = 4$:$L = \frac{4h}{2\pi} = \frac{2h}{\pi}$.

The difference between the radii of $n^{\text{th}}$ and $(n+1)^{\text{th}}$ orbits of a hydrogen atom is equal to the radius of the $(n-1)^{\text{th}}$ orbit of hydrogen. The angular momentum of the electron in the $n^{\text{th}}$ orbit is . . . . . . ($h$ is Planck's constant).

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