If a_0 is denoted as the Bohr radius of the h | vedclass

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

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$\frac{4 \pi a_0}{n}$

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(C) According to the Bohr quantization condition,the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $2 \pi r_n = n \lambda$ The radius of the $n^{th}$ orbit is given by $r_n = n^2 a_0$. Substituting $r_n$ into the equation: $2 \pi (n^2 a_0) = n \lambda$ Solving for $\lambda$: $\lambda = \frac{2 \pi n^2 a_0}{n} = 2 \pi n a_0$ Note: The original question asked for the second orbit $(n=2)$,which would yield $\lambda = 4 \pi a_0$. Given the options provided,the expression for the $n^{th}$ orbit is derived as $2 \pi n a_0$.

If $a_0$ is denoted as the Bohr radius of the hydrogen atom,then what is the de-Broglie wavelength $(\lambda)$ of the electron present in the $n^{th}$ orbit of the hydrogen atom?

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