A hydrogen atom is in its n^{text{th}} energy | vedclass

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(D) The angular momentum of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom is given by Bohr's quantization condition:$L = mvr = \frac{nh}

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$\lambda \propto n$

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(D) The angular momentum of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom is given by Bohr's quantization condition:$L = mvr = \frac{nh}{2\pi}$From the de-Broglie wavelength formula,we have $\lambda = \frac{h}{mv}$,which implies $\frac{mv}{h} = \frac{1}{\lambda}$.Substituting this into the angular momentum equation:$r \cdot \frac{1}{\lambda} = \frac{n}{2\pi} \Rightarrow \lambda = \frac{2\pi r}{n}$We know that the radius of the $n^{	ext{th}}$ orbit is proportional to $n^2$ $(r \propto n^2)$.Substituting $r \propto n^2$ into the expression for $\lambda$:$\lambda \propto \frac{n^2}{n} \Rightarrow \lambda \propto n$.

$A$ hydrogen atom is in its $n^{\text{th}}$ energy state. If the de-Broglie wavelength of the electron is $\lambda$,then:

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