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(D) For a hydrogen atom, the radius of the $n^{th}$ orbit is given by $r_n = n^2 a_0$. For the third excited state, the principal quantum number $n =

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$8$

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(D) For a hydrogen atom, the radius of the $n^{th}$ orbit is given by $r_n = n^2 a_0$. For the third excited state, the principal quantum number $n = 4$ (since ground state is $n=1$, first excited is $n=2$, second is $n=3$, and third is $n=4$). The radius of the fourth orbit is $r_4 = 4^2 a_0 = 16 a_0$. According to Bohr's postulate, the circumference of the orbit is an integral multiple of the de Broglie wavelength: $2\pi r_n = n \lambda$. Substituting the values for $n=4$: $2\pi (16 a_0) = 4 \lambda$. Solving for $\lambda$: $\lambda = \frac{32\pi a_0}{4} = 8\pi a_0$.

If the radius of the first Bohr orbit of a hydrogen atom is $a_0$, what will be the de Broglie wavelength of an electron in its third excited state (in $\pi a_0$)?

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