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(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$, where $a_0$ is the radius of the first orbit.For the second orbit $(n = 2)$, th

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$4$

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(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$, where $a_0$ is the radius of the first orbit.For the second orbit $(n = 2)$, the radius $r_2 = a_0 	imes (2)^2 = 4 a_0$.According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $n \lambda = 2 \pi r_n$.For $n = 2$, we have $2 \lambda = 2 \pi r_2$.Substituting $r_2 = 4 a_0$, we get $2 \lambda = 2 \pi (4 a_0)$.Therefore, $\lambda = 4 \pi a_0$.

If $a_0$ is the radius of the first Bohr's orbit of the $H$-atom, the de-Broglie wavelength of an electron revolving in the second Bohr's orbit will be: (in $\pi a_0$)

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