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(A) According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de Broglie wavelength: $2 \pi r = n \l

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$8$

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(A) According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de Broglie wavelength: $2 \pi r = n \lambda$. For the $n^{th}$ Bohr orbit, the radius is given by $r = n^{2} a_{0}$, where $a_{0}$ is the Bohr radius. For the $4^{th}$ orbit $(n = 4)$, the radius is $r = 4^{2} a_{0} = 16 a_{0}$. Substituting these values into the quantization equation: $2 \pi (16 a_{0}) = 4 \lambda$. Solving for $\lambda$: $\lambda = \frac{32 \pi a_{0}}{4} = 8 \pi a_{0}$.

The de Broglie wavelength of an electron in the $4^{th}$ Bohr orbit is (in $\pi a_{0}$)

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